Question

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?

Answer 1

Given: The mass of the metal block is 0.20 kg, the initial temperature of the metal is 150 °C, final temperature is 40 °C, calorimeter has water equivalent of mass is 0.025 kg, the density of the water is 1  kg/ m 3 and volume of water is 150  cm 3 .

The heat lost by the metal is given as,

ΔQ=mcΔT =mc( T 1 − T 2 )

Where, the heat energy of the metal is ΔQ, mass of the metal is m, specific heat of the metal is c, the initial temperature of the metal is T 1 and final temperature of the metal is T 2 .

By substituting the given values in the above expression, we get

ΔQ=0.20× 10 3 ×c×( 150−40 ) ΔQ=( 22× 10 3 c ) cal

The heat gained by the water and calorimeter is given as,

Δ Q ′ = m 1 c ′ Δ T ′ =( M+m' ) c ′ Δ T ′

Where, change in temperature of water is Δ T ′ , the mass of water at temperature 27 °C is M, the calorimeter has water equivalent of mass is m ′ , specific heat of water is c ′ and the heat gained by water is Δ Q ′ .

By substituting the given values in the above expression, we get

Δ Q ′ =( 150+25 )×4.186×( 40−27 ) =9523.15 cal

Now the heat lost by the metal is equal to the heat gained by the water and colorimeter system.

The equilibrium of the heat energy is given as,

ΔQ=Δ Q ′

By substituting the values in the above expression, we get

( 22× 10 3 c )=9523.15 c=0.43  Jg −1 K −1

Thus, the specific heat of metal is 0.43  Jg −1 K −1 .

If some heat is lost to surrounding then the value of c will be smaller than the actual value.