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. Use the following data to answer the question below:
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
Given the data above, determine the total heat energy gained by the container and water.

Answer 1

Answer:

$Q_a=330 cal$

$Q_w=6000cal$

Explanation:

From the question we are told that

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

a) Heat energy on container

Generally the formula for mathematically solving heat gain

      $Q_c=M_c *C_c*( \triangle T)$

Therefore imputing variables we have

      $Q_a=50g *0.22*50-20$  

      $Q_a=330 cal$

b) Heat energy on water

Generally the formula for mathematically solving heat gain

       $Q_w=M_w *C_w*( \triangle T)$

Therefore imputing variables we have

       $Q_w=200 *1* 50-204$

       $Q_w=6000cal$