Experiment 6: The specific heat of nickel is measured using a simple experimenta

Question

Experiment 6: The specific heat of nickel is measured using a simple experimental setup. Besides a sample of nickel, all that is needed for this experiment are hot as well as cold water, Styrofoam cups and a thermometer. 1.The Styrofoam cups function as calorimeter, and before the heat capacity of nickel is measured, the heat capacity of the calorimeter needs to be determined. Hot water and cold water are mixed in the Styrofoam cups, and the change in temperature is measured. From the following data, calculate the heat capacity C_cal of the calorimeter: 2. Having established a value for C_cal, the same calorimeter is used to determine the specific heat of nickel. To do so, a 78.8 g sample of nickel is heated to 99.2 C and transferred to 99.2 mL of water at 26.1 C. The final temperature of the water and metal is found to be 32.1 C. What is the specific heat S_NI of nickel? 3. Finally, the accuracy of the thermometer is checked measuring the boiling point and the freezing point of water. The thermometer reads 102.0 C when water is boiling and 2.0 C at the freezing point of water. Although the thermometer is clearly not calibrated, explain why this will not affect the results of the experiment.

Explanation / Answer

For measurement of specific heat of nickel we have to find out the heat capacity of calorimeter first.

Experiment 1

The heat given by hot water will be absorbed by cold water and calorimeter( foam cups)

The heat given by hot water = Mass of hot water x specific heat of water X change in temperature = Qh

Mass of hot water = 99.1- 74.6 = 24.5 grams

Change in temperature = 99.5 - 40.1 = 59.4 0C

Qh = 24. 5 X 4.18 X 59.4 = 6083.154 Joules

Heat absrobed by cold water= mass of cold water X speific heat of water X change in temperature = Qc

Mass of cold water = 74.6 - 6.2 = 68.4 grams

Change in temperature = 40.1 - 24 = 16.1 0C

Qc = 68.4 X 4.18 X 16.1 = 4603.183 Joules

Heat absrobed by calorimeter= Heat capacity X change in temperature = Qcal = Heat capacity X (40.1 - 24)

Qh = Qc + Qcal

6083.154 = 4603.183 Joules + Heat capacity X 16.1 0C

Heat capacity = 91.92 J / 0C

Experiment 2:

Heat given by hot metal nickel will be absrobed by water and calorimeter

Heat given by nickel = Mass of nickel X specific heat of nickel X change in temperature= QNi

QNi = 78.8grams X Speicifi heat X (99.2 - 32.1)

Heat absrobed by water = Mass of water X specific heat of nickel X change in temperature= Qw

Qw = 99.2 grams X 4.18 X (32.1 - 26.1) = 2487.936 Joules

Heat absorbed by calorimeter = Heat capacity X change in temperature = Qcal

Qcal = 91.92 X (32.1 - 26.1) = 551.52 Joules

QNi = 2487.936 Joules + 551.52 Joules = 78.8grams X Speicific heat X (99.2 - 32.1)

Specific heat of nickel = 3039.456 / 78.8 X 6 =0.575 J / g 0C

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