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 261g
Mass of the steam 11g
Specific heat of aluminum 0.22 cal/g°C

Heat energy gained by the container

a. 770 cal
b. 550 cal
c. 330 cal
d. 220 cal

Answer 1

Answer:

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.

Answer 2

Answer:

330 cal

Explanation: