23. What mass of water will give up 240 calories
when its temperature drops from 80°C to 68°C?

20g

Explanation:

Given parameters:

Quantity of heat  = 240calories

Initial temperature  = 80°C

Final temperature  = 68°C

Unknown:

Mass of water = ?

Solution;

To solve this problem, we use the expression:

         Q  = m c (t₂  - t₁)

Q is the quantity of heat

m is the mass

c is the specific heat = 1 cal/g°C

t₂ is the final temperature

t₁ is the initial temperature

  Insert the parameters and solve;

  Change in temperature is an absolute value

          240 = m x 1 x (68 - 80)

           m  = 20g