Question

You heat 18.57 g of a solid in a test tube to 99.8°C and then add the solid to 65.50 g of water in a coffee-cup calorimeter. The water temperature changes from 24.46°C to 29.27°C. Find the specific heat capacity of the solid in J/g°C. The specific heat capacity of water is 4.184 J/g°C. Enter to 3 decimal places.

Answer 1

Ans. Amount of gained by water when its temperature increases from 24.46⁰C to 29.27⁰C is given by-

            q = m s dT                            - equation 1

Where,

q = heat absorbed

m = amount of water

s = specific heat of water [ 4.184 J g^-10C^-1]

dT = Final temperature – Initial temperature

= 29.27⁰C – (24.46⁰C) = 4.81⁰C

Putting the values in equation 1-

            q = 65.50 g x (4.184 J g^-10C^-1) x 4.81⁰C = 1318.19012 J

# Amount of heat gained by water must be equal to the amount of heat lost by the solid.

So,

            Amount of heat lost by solid, q1 = - 1318.19012 J (the –ve gin indicates loss of heat)

When placed in calorimeter, the solid attains thermal equilibrium with the water. So, the final temperature of the solid = 29.27⁰C

            dT = 29.27⁰C – 99.8⁰C = -70.53⁰C ; the –ve sign indicates reduction in temperature

Again, using equation 1 for the solid-

            - 1318.19012 J = 18.57 g x s x (-70.53⁰C)

            Or, s = 1318.19012 J / (18.57 g x 70.53⁰C) = 1.006 J g^-10C^-1

Hence, specific heat (specific heat capacity) of the solid = 1.006 J g^-10C^-1