Question

Calculate the amount of heat required to convert 40g of ice at -100C to water at 500C.

(specific heat capacity of ice = 2100 J/kg K, specific latent heat of fusion= 336000 J/kg,

specific heat capacity of water = 4200J/kgK)​

Answer 1

Explanation:

Let final temperature of water when all the ice has melted = T°C.

Amount of heat lost when 200g of water at 50°C cools to T°C =200 ×4.2 ×(50 − T) = 42000 − 840T

Amount of heat gained when 40g of ice at 0oC converts into water at 0°C.= 40 ×336J = 13440 J

Amount of heat gained when temperature of 40g of water at 0°C rises to T°C = 40 ×4.2 ×(T-0) = 168T

We know thatAmount of heat gained = amount of heat energy lost.

13440 + 168T = 42000 − 840T

= 42000 − 13440 = 1008T

T = 28560/1008 = 28.33° C

Hope it helps ☺☺