Question

An investigation has been completed, similar to the one on latent heat of fusion, where steam is bubbled through a container of water. Steam condenses and the lost energy heats the water and container. Given the following data, do the calculations requested.

Mass of the aluminum container: 50 g
Mass of the container and water: 250 g
Mass of the water: 200 g
Initial temperature of the container and water: 20°C
Temperature of the steam: 100°C
Final temperature of the container, water, and condensed steam: 50°C
Mass of the container, water, and condensed steam: 261 g
Mass of the steam: 11 g
Specific heat of aluminum: 0.22 cal/g°C
Heat energy gained by the water:

A. 6,000 cal
B. 4,000 cal
C. 14,000 cal
D. 10,000 cal

Answer 1

The heat energy gained by water is 62.8 kJ. The heat energy gained by the aluminum container is 27.0 kJ. The percentage of heat going into the water is 70%.

Step-by-step explanation:

To calculate the heat transferred to the water, we can use the formula:

Qw = mwcwAT

where Qw is the heat transferred to the water, mw is the mass of the water, cw is the specific heat of water, and AT is the change in temperature. Substituting the given values:

Qw = (0.250 kg)(4186 J/kg°C)(60.0°C) = 62.8 kJ

To calculate the heat transferred to the aluminum, we can use the formula:

Qa = MAlCaAT

where Qa is the heat transferred to the aluminum, MAl is the mass of the aluminum container, Ca is the specific heat of aluminum, and AT is the change in temperature. Substituting the given values:

Qa = (0.500 kg)(900 J/kg°C)(60.0°C) = 27.0 kJ

To find the total transferred heat, we can add the heat transferred to the water and the heat transferred to the aluminum:

Total transferred heat = Qw + Qa = 62.8 kJ + 27.0 kJ = 89.8 kJ

To calculate the percentage of heat going into the water, we can use the formula:

Percentage of heat into water = (Qw / Total transferred heat) x 100%

Substituting the given values:

Percentage of heat into water = (62.8 kJ / 89.8 kJ) x 100% = 70%