PHYSpedia - User contributions [en]

Yeti

2016-11-18T18:24:43Z

Bchrisler: /* Newton's Law of Cooling */

=Thermal Physics - 2016 Fall=
==Introduction==
In Fall 2016 semester, the Thermal Physic class ran an experiment to measure the performance of a Yeti 20 ounce insulated coffee cup. These Yeti cups had become very popular over the previous year, most people that had one said that it was unbelievable how long they would keep coffee hot, so we wanted to find out just how long that was.
==Procedure==

===Materials used===

:Vernier's original ''LabQuest'' (multimeter), that we ran in the computer with the Logger Pro softrware Logger Pro 3.5.8.1

:Insulated cups of the brand ''YETI''

:Standard Styrofoam coffee cups

:Insulated coozie of the brand ''YETI''

:Water heater

:Marker

:Digital Scale

----

The first step was to make the appropiate setting in order to have everything ready to not lose much time on pouring the hot water into the cups, so that the temperature of the water
didn't decrease much when losing its heat to the surrounding environment. The previous setting to pouring the water in the cups consisted in starting the Vernier LabQuest, plugging the temperature
probes in the correspondant plugg-ins. We also opened the Logger Pro 3.8.5.1 programm in the computer and made sure that it would record the data for a long enough amount of time for the water to decrease
to a temperature of 40 Celsius degrees.

It was important to take into consideration the initial temperature of the surrounding environment, so we made sure to write down the room temperature before the collection of data started running.
The temperature of the room was also measured after the data collection was done, in order to have the most accurate room temperature to define better how well insulated the ''YETI'' cups actually are.

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-T_a)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>T_a</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. <math>\alpha</math> is dependent on the surface area

<math>\alpha = h/A</math>

===Derivation===

Solve equation for temperature as a function of time

<math> - \frac {h}{mc} \int\limits_{0}^{t}\, dt = \int\limits_{T_a}^{T(t)}\frac {dT}{T-T_a}</math>

<math>-\frac {ht}{mc} + C = ln(T(t)-T_a))</math>

<math>e^{-\frac {ht}{mc}} + C = T(t)-Ta</math>

<math>Ce^{-\frac {ht}{mc}} + T_a = T(t)</math>

Solving for C using initial conditions

<math>Ce^{0} + T_a = T(0)</math>

<math>C = T(0) - T_a</math>

Substituting C

<math>(T_o - T_a)e^{-\frac {ht}{mc}} + T_a = T(t)</math>

Solving for h

<math>(T_o - T_a)e^{-\frac {ht}{mc}} = T(t) - Ta</math>

<math>e^{-\frac {ht}{mc}} = \frac {T(t) - T_a}{T_o - T_a}</math>

<math>-\frac {ht}{mc} = \ln \left(\frac {T(t) - T_a}{T_o - T_a}\right)</math>

<math>-\frac {ht}{mc} = \ln \left(\frac {T(t) - T_a}{T_o - T_a}\right)</math>

<math>h =- \frac{mc}{t} \ln \left(\frac {T(t) - T_a}{T_o - T_a}\right)</math>

Yeti

2016-11-18T18:20:40Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-18T18:20:16Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-18T18:18:15Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-18T18:17:47Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T18:17:30Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-18T18:16:59Z

Bchrisler: /* Analysis */

Yeti

2016-11-18T18:16:19Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T18:15:44Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T18:11:46Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T18:10:17Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T18:08:33Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T18:07:13Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T18:03:12Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T17:59:44Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T17:58:48Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-18T17:58:26Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T17:57:33Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-18T17:57:08Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-18T17:56:48Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T17:56:30Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T17:55:07Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T17:54:11Z

Bchrisler: /* Derivation */

Yeti

2016-11-18T17:53:34Z

Bchrisler: /* Derivation */

=Thermal Physics - 2016 Fall=
==Introduction==
In Fall 2016 semester, the Thermal Physic class ran an experiment to measure the performance of a Yeti 20 ounce insulated coffee cup. These Yeti cups had become very popular over the previous year, most people that had one said that it was unbelievable how long they would keep coffee hot, so we wanted to find out just how long that was.
==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>\alpha = h/a</math>

====Derivation====

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(0) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>h = \frac{-mc}{t} ln (\frac {T(t) - Ta}{To - Ta})</math>

Yeti

2016-11-18T17:51:34Z

Bchrisler: /* Derivation */

=Thermal Physics - 2016 Fall=
==Introduction==
In Fall 2016 semester, the Thermal Physic class ran an experiment to measure the performance of a Yeti 20 ounce insulated coffee cup. These Yeti cups had become very popular over the previous year, most people that had one said that it was unbelievable how long they would keep coffee hot, so we wanted to find out just how long that was.
==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>\alpha = h/a</math>

====Derivation====

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^0 + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(0) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>h = \frac{-mc}{t} ln (\frac {T(t) - Ta}{To - Ta})</math>

Yeti

2016-11-18T17:50:46Z

Bchrisler: /* Derivation */

=Thermal Physics - 2016 Fall=
==Introduction==
In Fall 2016 semester, the Thermal Physic class ran an experiment to measure the performance of a Yeti 20 ounce insulated coffee cup. These Yeti cups had become very popular over the previous year, most people that had one said that it was unbelievable how long they would keep coffee hot, so we wanted to find out just how long that was.
==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>\alpha = h/a</math>

====Derivation====

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(0) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>h = \frac{-mc}{t} ln (\frac {T(t) - Ta}{To - Ta})</math>

Yeti

2016-11-18T17:49:17Z

Bchrisler: /* Derivation */

=Thermal Physics - 2016 Fall=
==Introduction==
In Fall 2016 semester, the Thermal Physic class ran an experiment to measure the performance of a Yeti 20 ounce insulated coffee cup. These Yeti cups had become very popular over the previous year, most people that had one said that it was unbelievable how long they would keep coffee hot, so we wanted to find out just how long that was.
==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>\alpha = h/a</math>

====Derivation====

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^(-ht/mc) + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^(-ht/mc) = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>h = \frac{-mc}{t} ln (\frac {T(t) - Ta}{To - Ta})</math>

Yeti

2016-11-18T17:48:49Z

Bchrisler: /* Derivation */

=Thermal Physics - 2016 Fall=
==Introduction==
In Fall 2016 semester, the Thermal Physic class ran an experiment to measure the performance of a Yeti 20 ounce insulated coffee cup. These Yeti cups had become very popular over the previous year, most people that had one said that it was unbelievable how long they would keep coffee hot, so we wanted to find out just how long that was.
==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>\alpha = h/a</math>

====Derivation====

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^(-ht/mc) + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^(-ht/mc) = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^(-ht/mc) + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^(-ht/mc) = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>h = \frac{-mc}{t} ln (\frac {T(t) - Ta}{To - Ta})</math>

Yeti

2016-11-18T17:48:03Z

Bchrisler: /* Newton's Law of Cooling */

=Thermal Physics - 2016 Fall=
==Introduction==
In Fall 2016 semester, the Thermal Physic class ran an experiment to measure the performance of a Yeti 20 ounce insulated coffee cup. These Yeti cups had become very popular over the previous year, most people that had one said that it was unbelievable how long they would keep coffee hot, so we wanted to find out just how long that was.
==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>\alpha = h/a</math>

====Derivation====

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^(-ht/mc) + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^(-ht/mc) = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^(-ht/mc) + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^(-ht/mc) = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>h = \frac{-mc}{t} ln (\frac {T(t) - Ta}{To - Ta})</math>

Yeti

2016-11-18T17:46:27Z

Bchrisler: /* Derivation */

=Thermal Physics - 2016 Fall=
==Introduction==
In Fall 2016 semester, the Thermal Physic class ran an experiment to measure the performance of a Yeti 20 ounce insulated coffee cup. These Yeti cups had become very popular over the previous year, most people that had one said that it was unbelievable how long they would keep coffee hot, so we wanted to find out just how long that was.
==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>\alpha = h/a</math>

====Derivation====

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^(-ht/mc) + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^(-ht/mc) = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^(-ht/mc) + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^(-ht/mc) = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>h = -\frac {mc ln (\frac {T(t) - Ta}{To - Ta})t}</math>

Yeti

2016-11-18T17:45:54Z

Bchrisler: /* Derivation */

=Thermal Physics - 2016 Fall=
==Introduction==
In Fall 2016 semester, the Thermal Physic class ran an experiment to measure the performance of a Yeti 20 ounce insulated coffee cup. These Yeti cups had become very popular over the previous year, most people that had one said that it was unbelievable how long they would keep coffee hot, so we wanted to find out just how long that was.
==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>\alpha = h/a</math>

====Derivation====

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^{-ht/mc} + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>Ce^{h(0)} + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^{-ht/mc} + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^{-ht/mc} = T(t) - Ta</math>

<math>e^{-ht/mc} = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^(-ht/mc) + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^(-ht/mc) = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

Solving for the heat transfer coefficient

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>e^{-ht/mc} + C = T(t)-Ta</math>

<math>Ce^(-ht/mc) + Ta = T(t)</math>

Solving for C using initial conditions

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>Ce^(h(0)) + Ta = T(0)</math>

<math>C = T(0) - Ta</math>

Substituting C

<math>(To - Ta)e^(-ht/mc) + Ta = T(t)</math>

Solving for h

<math>(To - Ta)e^(-ht/mc) = T(t) - Ta</math>

<math>e^(-ht/mc) = \frac {T(t) - Ta}{To - Ta}</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>-ht/mc = ln (\frac {T(t) - Ta}{To - Ta})</math>

<math>h = -\frac {mc ln (\frac {T(t) - Ta}{To - Ta})}/t</math>

Yeti

2016-11-18T17:44:45Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-18T17:43:34Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-18T17:36:03Z

Bchrisler: /* Newton's Law of Cooling */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>\alpha = h/a</math>

====Derivation====

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>-e^(ht/mc) + C = T(t)-Ta</math>

<math>-Ce^(ht) + Ta = T(t)</math>

Yeti

2016-11-18T17:35:03Z

Bchrisler: /* Analysis */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

<math>/alpha = h/a</math>

====Derivation====

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>-e^(ht/mc) + C = T(t)-Ta</math>

<math>-Ce^(ht) + Ta = T(t)</math>

Yeti

2016-11-18T17:33:57Z

Bchrisler: /* Newton's Law of Cooling */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

/alpha = h/a

====Derivations====

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>-e^(ht/mc) + C = T(t)-Ta</math>

<math>-Ce^(ht) + Ta = T(t)</math>

Yeti

2016-11-18T17:33:31Z

Bchrisler: /* Analysis */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

/alpha = h/a

====Newton's Law of Cooling====

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>-e^(ht/mc) + C = T(t)-Ta</math>

<math>-Ce^(ht) + Ta = T(t)</math>

Yeti

2016-11-18T17:32:55Z

Bchrisler: /* Newton's Law of Cooling */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

<math>Q</math> is thermal energy

<math>h</math> is the heat transfer coefficient

<math>T</math> is the temperature of the water

<math>Ta</math> is the temperature of the air (environment)

<math>t</math> is time

<math>m</math> is mass of water

<math>c</math> is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

/alpha = h/a

<math>-h/mc dt = dT/(T-Ta)</math>

<math>-ht/mc + C = ln(T(t)-Ta))</math>

<math>-e^(ht/mc) + C = T(t)-Ta</math>

<math>-Ce^(ht) + Ta = T(t)</math>

Yeti

2016-11-07T18:18:12Z

Bchrisler: /* Newton's Law of Cooling */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===

<math>dQ = -h(T-Ta)dt = mcdT</math>

'''Q''' is thermal energy

'''h''' is the heat transfer coefficient

'''T''' is the temperature of the water

'''Ta''' is the temperature of the air (environment)

'''t''' is time

'''m''' is mass of water

'''c''' is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

/alpha = h/a

-h/mc dt = dT/(T-Ta)

-ht/mc + C = ln(T(t)-Ta))

-e^(ht/mc) + C = T(t)-Ta

-Ce^(ht) + Ta = T(t)

Yeti

2016-11-07T18:09:11Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-07T18:08:10Z

Bchrisler: /* Newton's Law of Cooling */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===
dQ = -h(T-Ta)dt = mcdT

'''Q''' is thermal energy

'''h''' is the heat transfer coefficient

'''T''' is the temperature of the water

'''Ta''' is the temperature of the air (environment)

'''t''' is time

'''m''' is mass of water

'''c''' is specific heat capacity of water

The heat transfer coefficient '''h''' is depended on the material. Alpha is dependent on the surface area and beta is dependent on

/alpha = h/a

Yeti

2016-11-07T18:06:12Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-07T18:04:15Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-07T18:01:20Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-07T18:01:00Z

Bchrisler: /* Newton's Law of Cooling */

Yeti

2016-11-07T17:58:21Z

Bchrisler: /* Newton's Law of Cooling */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===
<math>3</math>

'''Q''' is thermal energy

'''h''' is the heat transfer coefficient

'''T''' is the temperature of the water

'''Ta''' is the temperature of the air (environment)

'''t''' is time

'''m''' is mass of water

'''c''' is specific heat capacity of water

Yeti

2016-11-07T17:57:45Z

Bchrisler: /* Newton's Law of Cooling */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==
[[File:Lydia.png]]

==Analysis==
===Newton's Law of Cooling===
<math>3</math>

Q is thermal energy

h is the heat transfer coefficient

T is the temperature of the water

Ta is the temperature of the air (environment)

t is time

m is mass of water

c is specific heat capacity of water

Yeti

2016-11-07T17:55:28Z

Bchrisler: /* Derivation */

=Thermal Physics - 2016 Fall=
==Introduction==

==Procedure==

==Data==

==Analysis==
===Newton's Law of Cooling===
<math>3</math>

Yeti

2016-11-07T17:47:40Z

Bchrisler: /* Analysis */

=Thermal Physics - 2016 Fall=

==Procedure==

==Data==

==Analysis==
===Derivation===
<math>3</math>