Question

a block of mass 2.5kg is heated to temperature of500°c and placed in a large ice block. what is the maximum amount of ice that can melt specific heat loss=0.1cal/g°c​

Answer 1

Let ice of mass, m is melted.

you know as well, ice is at 0°C and given block of mass 2.5 kg at 500°C .

so, heat lost by block = heat gained by ice.

heat lost by block, Q = MS∆T

here M is mass of block, S is specific heat capacity of block and ∆T is change in temperature.

so, M = 2.5 kg = 2500g, S = 0.1Cal/g.°C

and ∆T = (500°C - 0°) = 500°C

so, Q = 2500 × 0.1 × 500 = 125000 cal

latent heat of fusion , Lf = 80cal/g

so, heat gained by ice = mLf

= m × 80

now, 125000 = m × 80

or, m = 125000/80 = 12500/8 = 1562.5g = 1.5625kg

hence, 1.5625 kg of ice is melted.