Question

A 45.2 g sample of Silver at 101.0 ºC is dropped into 300.0 g of water at 30.6 ºC. What is the final temperature of the water?

Answer 1

Final temperature of water = 31.2° C

Step-by-step explanation:

Let the final temperature of water be T degrees

Concept: By energy conservation law

Heat lost by the silver = heat gained by water

Heat lost be silver Q1 = m1c1ΔT1

Q1 = 45.2/1000Kg×2.36J/KgK×(101-T)

Heat gained by water Q2 = m2c2ΔT2

Q2 = 300/1000Kg×4.186J/KgK(T-30.6)

Since, Q1=Q2

⇒ 45.2/1000Kg×2.36J/KgK×(101-T) = 300/1000Kg×4.186J/KgK(T-30.6)

Solving we get T = 31.2° C

Therefore, final temperature of water = 31.2° C