Q1 A 10kg solid at 100°C with a specific heat of 0.8 kj/kg°C is immersed in 40 kg of 20°C liquid with a specific heat of 4.0 kj/kg°C estimate the temperature after a long time if the Container is insulated (specific heat of water = 4.18 kj/kg).

(A) 30°C

(B) 28°C

(D) 24°C

Ans: (D) 24°C

Solution:

*Q=mcΔT

Q=Heat addition

m=mass

c= specific heat

ΔT= change in temperature

Heat given by solid = Heat taken by water

10×0.8×(100-T)= 40×4×(T- 20)

800 - 8T = 160T - 3200

4000 = 168T

T = 23.8°C = 24°C

Comments

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