Question

In an experiment, 25.5 g of metal was heated to 98.0°C and then quickly transferred to 150.0 g of water in a calorimeter. The initial temperature of the water was 23.0°C, and the final temperature after the addition of the metal was 32.5°C. Assume the calorimeter behaves ideally and does not absorb or release heat.

1.) What is the value of the specific heat capacity (in J/g•°C) of the metal?

_____ J/g•°C

Answer 1

Ans. Amount of heat changed (gained or lost) during attaining thermal equilibrium is given by-

            q = m s dT                            - equation 1

Where,

q = heat gained or lost

m = mass

s = specific heat

dT = Final temperature – Initial temperature

# At thermal equilibrium, total heat lost by hot metal (immersed in water bath) is equal to the amount of heat gained by water in calorimeter. To depict heat loss by meat, a –ve sign is used.

So,

            - q1 (metal) = q2 (water)

            Or, - 25.5 g x s₁ x (32.5 – 98.0)⁰C = 150.0 g x (4.184 J g^-10C^-1) x (32.5 – 23.0)⁰C

            Or, 1670.25 s₁ g ⁰C= 5962.2 J

            Or, s₁ = 5962.2 J / (1670.25 g ⁰C) = 3.57 J g^-10C^-1

Therefore, specific heat of metal, s₁ = 3.57 J g^-10C^-1