Measurement of the specific heat of aluminum in the 45°C - 65°C temperature range.In this part, you can use the following equipment ONLY:(1) A plastic cup with a cap which has a hole for the thermometer. (2) A digital thermometer with accuracy of 0. 1°C. The temperature of the sensor (10) is shown

Question

Measurement of the specific heat of aluminum in the 45°C - 65°C temperature range.

In this part, you can use the following equipment ONLY:

(1) A plastic cup with a cap which has a hole for the thermometer.
(2) A digital thermometer with accuracy of 0. 1°C. The temperature of the sensor (10) is shown on the top display, and bottom display shows the temperature of the room. Do not press max/min button: pressing max/min button changes the readings between current, maximal and minimal values. If water temperature exceeds 70°C, the thermometer shows "H" denoting it is out of its range.

Warning: Do Not Use the Thermometer to Measure the Temperature of Liquid Nitrogen! Thermometer Can Be Used Only in Part 1.
(3) An aluminum cylinder with a hole.
(4) Electronic scales with accuracy of 1 g. Make sure that the scales are situated on the flat surface. The button Tare/Zero serves as On/Off and sets the zero reading of the scales. Do not press any other buttons. Note: the scales automatically turn OFF after some time; you have to turn them ON back and reset the zero reading of the scales.
(5) A digital timer. Pressing left button shifts the timer from Clock mode to Stopwatch mode. In Stopwatch mode the middle button serves as Stop/Start, and the right button serves as Reset.
(6) Dewar flask with hot water.
(7) A jar for the used water.
(8) Plotting (graphing) paper (2 pages).
(9) Pieces of thread.

The results of the measurement of specific heat of aluminum will be used in Part 2 of the experimental problem.
The specific heat of aluminum should be determined from the comparison of two experimental curves:
1) the cooling curve of hot water in a plastic cup without the aluminum cylinder (the first experiment) ;
2) the cooling curve of hot water in a plastic cup with aluminum cylinder immersed (the second experiment) .
The specific heat of water is given [latex]c_{\mathrm{w}}[/latex] = 4.20 kJ/(kg K).
Density of water [latex]\rho_{\mathrm{w}}[/latex] = 1.00 • 10³ kg/m³ .
Density of aluminum[latex]\rho_{\mathrm{AI}}[/latex]= 2.70 • 10³ kg/m³ .
Warning: Be very careful with hot water. Remember that water at temperature T> 50°C can cause burns. Do Not Use Liquiid Nitrogen in This Part!

The task
(1 a) Derive theoretically an expression for aluminum specific heat [latex]c_{\mathrm{AI}}[/latex] in terms of experimentally measured quantities: mass [latex]m_{\mathrm{1}}[/latex] of hot water in the first experiment, mass [latex]m_{\mathrm{2}}[/latex] of hot water in the second experiment, mass m of aluminum cylinder and the ratio of heat capacities[latex]K={\frac{C_{1}}{C_{2}}},[/latex]where [latex]C_{\mathrm{1}}[/latex] is the heat capacity of water in the first experiment, [latex]C_{\mathrm{2}}[/latex] is the combined heat capacity of water and aluminum cylinder in the second experiment.

In parts (1b) and (1c) you will perform measurements to determine K. Parts (1b) and (1c) should be performed with closed caps. Assume that in this case heat exchange of the contents of the cup with the environment depends linearly on the difference in their temperatures. The linearity coefficient depends only on the level of the water in the cup. Make sure that the aluminum cylinder is fully immersed into water in part (1c). You can neglect the heat capacity of the cup.
( 1 b ) Perform the first experiment investigate the relationship between the temperature of water [latex]T_{\mathrm{1}}[/latex] and time t in the range of temperatures from 45°C to 65°C. Provide a table of measurements. Write the value of ml on the answer sheet.
(1c) Perform the second experiment investigate the relationship between the temperature of water with aluminum cylinder [latex]T_{\mathrm{2}}[/latex] and time t in the range of temperatures from 45°C to 65 °C. Provide a table of measurements. Write the values of m2 and m on the answer sheet.
(1 d) Use graphs to determine the ratio of the heat capacities[latex]K={\frac{C_{1}}{C_{\mathrm{2}}}}[/latex]and the uncertainty ΔK. Write the values of K and ΔK on the answer sheet.
(1e) Determine the numerical value of [latex]c_{\mathrm{Al}}[/latex] and estimate the uncertainty of measurement Δ[latex]c_{\mathrm{Al}}[/latex] Write the values of [latex]c_{\mathrm{Al}}[/latex] and Δ[latex]c_{\mathrm{Al}}[/latex] on the answer sheet.

Accepted Answer

(1a) The heat capacity of water in the first experiment is

[latex]C_{1}=c_{\mathrm{w}}m_{1}\,,[/latex] (1)

where [latex]c_{w}=4.2\,\ \mathrm{kj/}(\mathrm{kg}\;\mathrm{K})[/latex] is the specific heat of water and [latex]m_{\mathrm{1}}[/latex] is the mass of water in the experiment. The heat capacity of water with the aluminum cylinder immersed in it:

[latex]C_{2}=c_{\mathrm{w}}m_{2}+c_{\mathrm{Al}}m,[/latex] (2)

where [latex]m_{\mathrm{2}}[/latex] is the mass of water in the experiment, [latex]c_{\mathrm{Al}}[/latex] is the specific heat of aluminum and m is the mass of aluminum cylinder. The ratio of heat capacities is [latex]K={\frac{C_{1}}{C_{\mathrm{2}}}}[/latex]. Then, specific heat capacity of aluminum is determined by the formula

[latex]c_{\mathrm{Al}}={\frac{m_{1}-K\cdot m_{2}}{K\cdot m}}c_{\mathrm{w}}.[/latex] (3)

The ratio of heat capacities K can be determined from the experiment in (1 b) and (1c). To be able to extract K from these two experiments, one should perform the measurements in the regime such that the level of water is the same for both experiments. This can be done by marking the level of the water on the side of the cup by usual pen, or by choosing the masses of water such that the equation

[latex]{\frac{m_{1}}{ ho_{\mathrm{w}}}}={\frac{m_{2}}{ ho_{\mathrm{w}}}}+{\frac{m}{ ho_{\mathrm{AI}}}}[/latex]

is closely satisfied. Since Eq. (3) have the difference in the numerator, best results are obtained when ml is chosen to be close to m.

(1 b) The following table shows the [latex]T_{\mathrm{1}}[/latex] , - the temperature of hot water as it cools down, as a function of time t in the 45°C - 65°C temperature range:

Table--1--

The mass of water is [latex]m_{\mathrm{1}}[/latex] = (50 ± 1) g and the room temperature is [latex]T_{\mathrm{r}}[/latex] = (23, 4±0, 2)°C.

(1c) The following table shows the temperature [latex]T_{\mathrm{2}}[/latex] of hot water with aluminum cylinder immersed as the water cools down as a function of time t in the 45°C- 65°C temperature range:

Table--2--

The mass of aluminum cylinder is m = (69 ± 1) g, the mass of water is [latex]m_{\mathrm{2}}[/latex] = (27 ± 1) g, and the room temperature is [latex]T_{\mathrm{r}}[/latex] = (23, 4 ± 0, 2)°C.

The graphs [latex]T_{\mathrm{1}}[/latex] (t) and [latex]T_{\mathrm{2}}[/latex] (t) are shown below:

(1d) Water in the first experiment and water with aluminum cylinder immersed in the second experiment cool down because of heat exchange with the air in the room according to the following linear law:

[latex]\alpha(T-T_{r})d t=-\,C d T,[/latex] (4)

where ex is αconstant and C is heat capacity ([latex]C_{1}[/latex] or [latex]C_{2}[/latex]). If we integrate the expression (4), we get

[latex]T-T_{r}=A\mathrm{e}^{-{\frac{\mathrm{\alpha}}{c^{}}t}},[/latex] (5)

where A = [latex]T_{0}[/latex] - [latex]T_{r}[/latex]([latex]T_{0}[/latex] is the initial temperature of water in the experiment from the experimental data) .

Various methods can be used to obtain the ratio of heat capacities but the most precise result can be obtained from the following linear relation:

[latex]\ln(T-T_{r})\,=\,\ln A-{\frac{\alpha}{C}}t.[/latex] (6)

The graphs [latex]\mathsf{ln}\big[ T_{1}(t)-T_{r}\big][/latex] and [latex]\mathsf{ln}\big[ {-}T_{2}(t)-T_{r}\big][/latex] appear to be approximately linear and are shown below:

We can obtain the ratio [latex]K={\frac{C_{1}}{C_{2}}},[/latex] by comparing the slopes of the graphs derived from the first and second experiments. The value of K obtained in terms of the slopes of the two linear relationships is as follows:

[latex]K=\frac{b_{2}}{b_{1}}=\frac{-0.0439}{-0.0367}=1.196.[/latex]

And the uncertainty is:

[latex]\Delta K=K.\left({\frac{\Delta b_{1}}{b_{1}}}+{\frac{\Delta b_{2}}{b_{2}}} ight)=1.\;196\ .\left({\frac{0.\;0004}{0.\;0367}}+{\frac{0.\;005}{0.\;0439}} ight)[/latex]

[latex]=1.196 imes0.022=0.026,\quad\varepsilon_{K}=2\%.[/latex]

(1 e) At K = 1. 196 the specific heat of aluminum obtained from formula (3) is:

[latex]c_{\mathrm{Al}}=0.90\mathrm{~kJ/}(\mathrm{kg}.\mathrm{K}).[/latex]

And the uncertainty is:

[latex]\Delta c_{\mathrm{Al}}\,=\,c_{\mathrm{Al}}\Bigl({\frac{\Delta m_{1}+K\Delta m_{2}}{m_{1}-K m_{2}}}+{\frac{\Delta m}{m}}+{\frac{\Delta K}{K}}\,{\frac{m_{1}}{m_{1}-K m_{2}}}\Bigr)[/latex]

[latex]=0.90.\left({\frac{2.196\,~\mathrm{g}}{17.7\,{\mathrm{~g}}}}+{\frac{1\,{\mathrm{g}}}{69\,{\mathrm{g}}}}+{\frac{50\,{\mathrm{g}}}{17.7\,{\mathrm{g}}}{\frac{0.026}{10196}}} ight)[/latex].

= 0.90 · 0.2 = 0.18 kJ/(kg· K).

Question 7.EC.1: Measurement of the specific heat of aluminum in the 45°C - 6......

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