KnowledgeBoat Logo

Physics

A copper vessel of mass 100 g contains 150 g of water at 50 °C. How much ice is needed to cool it to 5 °C?

Given : Specific heat capacity of copper = 0.4 J g-1 ⁰C-1

Specific heat capacity of water = 4.2 J g-1 ⁰C-1

Specific latent heat of fusion of ice = 336 J g-1

Calorimetry

ICSE 2016

5 Likes

Answer

Heat energy imparted by vessel = 100 x 0.4 x (50 - 5) = 1800 J

Heat energy imparted by water = 150 x 4.2 x (50 - 5) = 28,350 J

Let m g of ice be used, then

Heat energy taken by ice to melt = m x 336 J

Heat energy taken by melted ice to raise it's temperature from 0 °C to 5 °C = m x 4.2 x (5 - 0) = 21m J

By law of conservation of energy

heat energy imparted by vessel and water = heat energy taken by ice and melted ice

1800 + 28,350 = 336m + 21m

357 m = 30,150

m = 30150357\dfrac{30150}{357} = 84.45 g

Answered By

3 Likes


Related Questions