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 20Â°C

Temperature of the steam 100Â°C

Final temperature of the container, water, and condensed steam 50Â°C

Mass of the container, water, and condensed steam 261g

Mass of the steam 11g

Specific heat of aluminum 0.22 cal/gÂ°C

Heat energy gained by the container

a. 770 cal

b. 550 cal

c. 330 cal

d. 220 cal

Question

Avah Perez

Chemistry

14 May, 03:51

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 20Â°C

Temperature of the steam 100Â°C

Final temperature of the container, water, and condensed steam 50Â°C

Mass of the container, water, and condensed steam 261g

Mass of the steam 11g

Specific heat of aluminum 0.22 cal/gÂ°C

Heat energy gained by the container

a. 770 cal

b. 550 cal

c. 330 cal

d. 220 cal

+4

Answers (1)

Know the Answer?

Claudia Merritt · Answer 1

c. 330 cal

Explanation:

Given the information from the question. We need to calculate heat energy gained by the container

The formula to be used is Q=M*C * (Change in T). Where Q = heat energy gained by the container, M = mass of aluminum, Change in T = Change in temperature and T = Specific heat of aluminum.

Now we have Q=M*C * (Change in T) = 50g*0.22 * (50-20) = 330 cal

The correct answer is option c. 330 cal.