Question

One way to experimentally measure the heat capacity of a Styrofoam cup calorimeter would be to melt a known mass of ice in warm water and measure the temperature change. Use the data below to determine the experimental heat capacity of a calorimeter. Use the literature heat of fusion for ice in your calculations. Assume the ice added is at 0.0 °C.

Data:

• Mass of ice added: 17.69 g

• Mass of water in calorimeter: 98.67 g

• Initial temperature of water: 28.7 °C

• Final temperature of water after melting ice: 12.9 °C

• Literature heat of fusion of ice: 6.02 kJ/mol

Question:

• Determine the experimental heat capacity of the calorimeter.

Answer 1

To find the calorimeter's heat capacity, calculate the heat lost by water, heat gained by ice, and use the difference to find calorimeter heat gain, considering the heat of fusion for ice and the specific heat capacity of water.

Step-by-step explanation:

To determine the experimental heat capacity of the calorimeter using a Styrofoam cup calorimeter, we first calculate the heat lost by the water using the formula:

• q = m * c * ΔT

Where:

• q is the heat lost by the water,
• m is the mass of the water (98.67 g),
• c is the specific heat capacity of water (4.18 J/g°C), and
• ΔT is the change in temperature (28.7 °C - 12.9 °C = 15.8 °C).

We then calculate the heat gained by the ice to melt it, and to warm the resulting water to the final temperature, using the heat of fusion for ice (334 J/g or the provided 6.02 kJ/mol) and the specific heat capacity of water. The heat capacity of the calorimeter (Ccal) is found from the difference between the heat lost by the water and the heat gained by the ice.

Assuming no heat loss to the surroundings and that the heat lost by the water is gained by the ice and the calorimeter, we have:

• Heat lost by water (qwater) = Heat gained by ice (qice) + Heat gained by calorimeter (qcal).

Using:

• qice = (mass of ice) * (heat of fusion) + (mass of ice meltwater) * c * (final temperature - initial temperature of ice).

Therefore:

• Ccal = (qwater - qice) / ΔTcalorimeter

Where ΔTcalorimeter is the change in temperature of the water in the calorimeter, which is equal to the final temperature of the water minus the initial temperature of the water. By inputting the known values into these equations, we can solve for the calorimeter's heat capacity.