Question

2. It is given that the specific heat of copper is one
tenth that of water. A copper ball of 100 g at 200°C
is put into 100 cc water at 24°C. Assuming no
heat loss to the surroundings. The final
temperature of the system is
(1) 60°C
(2) 55°C
(3) 40°C
(4) 35°C​

Answer 1

Answer:

option 3 is correct

Explanation:

given Cp of copper = 1/10 * Cp of water

let T be the final temperature

100 cc of water = 100 g

heat rejected by copper =

heat absorbed by water

100*(1/10)*(200-T) = 100*(T-24)

20 - (T/10) = T-24

44 = (11*T/10)

T = 44*10/11 = 40 ℃

Answer 2

Answer:

Explanation:

given Cp of copper = 1/10 * Cp of water

let T be the final temperature

100 cc of water = 100 g

heat rejected by copper =

heat absorbed by water

100*(1/10)*(200-T) = 100*(T-24)

20 - (T/10) = T-24

44 = (11*T/10)

T = 44*10/11 = 40 ℃