The specific heat capacity of an object A is X. It is kept in contact with another object B of twice the mass and specific heat capacity of twice of X. If the initial temperature of A is 100 ° C and B is 200 ° C, what will be change in the temperature of A?
The specific heat capacity of an object A is X. It is kept in contact with another object B of twice the mass and specific heat capacity of twice of X. If the initial temperature of A is 100 ° C and B is 200 ° C, what will be change in the temperature of A? Correct Answer 60 K
Concept:
- Specific Heat Capacity: Specific heat capacity is the energy required to increase the temperature of the material of a certain mass by 1°C.
- SI unit is J/ (kg·K).
- Heat gained by X = Heat lost by Y
H = m c ΔT, Where m = mass, c = specific heat capacity, ΔT = temperature change.
- When two objects are kept close, heat is transferred takes place from a hotter body to a cooler one.
- Finally, both objects reach the equilibrium temperature.
Calculation:
The specific heat of A = X
Mass of A = m (let)
Temperature of X = T1 = 100 °C = 373 K
specific heat of B = Y
Mass of B = 2 m ( twice of A )
Specific heat of A = 2 X (twice of X)
Temperature of Y = T2 = 200 ° C = 473 K
Let the equibrium temperature be T
Then
Heat lost by B = Heat gain by A
Heat lost of B = 2 m (2X) (T2 - T) = 4 m X (473 K - T)
Heat gain by A = m X (T - T2) = m X (T - 273K)
Equating
4 m X (473 - T) = m X (T - 273K)
⇒ 4 × 473 - 4 T = T - 273 K
⇒ -5 T = -273 - 4 × 473 = - 2165 K
⇒ T = 433 K
This is the equilibrium temperature.
The change in temperature = (T - T1) = 433 K - 373 K = 60 K
So, the change in temperature of X = 60 K.