Question

1. Calculate the heat which is associated with the change in temperature of the water (q_water):
2. The specific heat of aluminum is 0.897 J/gºC. Calculate the heat which is associated with the change in temperature of the aluminum (q_Al):
3. Answer the following:
4. 1. Write the equation that relates q_water, q_Al and q_calorimeterin an insulated system:
5. 2. Using your answers from Q2, Q3 and Q4a, calculate the heat associated with the calorimeter, q_calorimeter:
6. Calculate the calorimeter constant, C_calorimeter:

Chart for question 1-6

Mass of metal Al: 7.4760g

Volume of water: 100mL

Mass of water:

Initial temp of water: 25 celsius

Initial temp of metal: 200 celsius

Max temp of water and metal: 27.65 celsius

1. Using the calorimeter constant determined in Part I, determine the specific heat of lead (C_Pb). Be sure to use the right equation which takes into account the C_calorimeter, and show all work and equations used for full credit:

Chart for question 1

Mass of metal Pb: 33.3590g

Volume of water: 100mL

Mass of water:

Initial temp of water: 25 celsius

Initial temp of metal: 200 celsius

Max temp of water and metal: 26.73 celsius

Answer 1

Heat lost by Aluminum is gained by water inside the calorimeter and calorimeter.

The equation relation the heat exchange process is

q_Al = q_water + q_calorimeter

2)

Specific heat of Al = 0.897 J/g°C

Max temperature of Al and water = 27.65°C

Initial temperature of Al = 200°C

∆T(Al) = (200-27.65) = 172.35°C

Mass of Al = 7.4760 g

q_Al = mCp∆T = 7.4760(0.897) (172.35)

q_Al = 1155.7742 J

Specific heat of water = 4.184 J/g°C

Volume of water = 100 ml

Density of water = 1g/ml

Mass of water = 100(1) = 100 g

Initial temperature of water = 25°C

∆T(w) = (27.65-25) = 2.65°C

q_w = 100(4.184) (2.65) = 1108.76 J

q_Al = q_water + q_calorimeter

q_calorimeter = 1155.7742- 1108.76

q_calorimeter = 47.0142 J

q_calorimeter = C_calorimeter(∆T)

∆T(calorimeter) = 27.65-25 = 2.65°C

47.0142 = C_calorimeter (2.65)

C_calorimeter = 17.7412 J/°C

B)

Mass of lead = 33.3590 g

Initial temperature of lead = 200°C

Max temperature of water and metal = 26.73 °C

Initial temperature of water = 25°C

Volume of water = 100 mL

Density of water = 1 g/ml

Mass of water = 100(1) = 100 g

Specifc heat of water = 4.184 J/mol°C

∆T(water) = (26.73-25) = 1.73 °C

q_w = mCp∆T = 100(4.184) (1.73) = 723.832 J

q_calorimeter = C_calorimeter ∆T

q_calorimeter = 17.7412(1.73) = 30.6922 J

Doing heat balance

q_lead = q_calorimeter + q_{​​​​​​water}

33.3590(C_pb) (200-26.73) = 30.6922+ 723.832

C_pb = 0.13053 J/g°C

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