Question

An 80 g piece of ice is thrown into 140 g of water that is 50°C. Determine the final temperature of water once the ice has fully melted. (Assume there is no heat loss to the environment.) Specific heat capacity of water = 4.2 J g⁻¹ K⁻¹, Specific latent heat of fusion of ice = 336 J g⁻¹.

Answer 1

Given:

Mass of ice = 80 g

Mass of water = 140 g

Initial temperature of water = 50°C

Specific heat capacity of water, c = 4.2 J g⁻¹ K⁻¹

Latent heat of fusion of ice, L = 336 J g⁻¹

No heat loss

Let the final temperature be T°C.

1. Heat required to melt the ice:

Q₁ ​= mL

= 80 × 336

= 26880 J

2. Heat lost by the warm water:

The water cools from 50°C to T°C:

Q₂ ​= 140 × 4.2 (50 − T)

Since heat lost = heat gained,

140 × 4.2 (50 − T) = 26880 + 80 × 4.2 T

588 (50 − T) = 26880 + 336 T

29400 − 588 T = 26880 + 336 T

2520 = 924 T

T = 2.73°C

So, the final temperature of the water is approximately 2.7°C.