Question

A metal ball of mass 400 g and at 600°C is placed on a block of ice, when 428 g of ice melts. If the sp. heat capacity of the metal ball is 0.6 Jg⁻¹°C⁻¹, calculate sp. latent heat of ice.

Answer 1

As the block does not melt completely, therefore, final temperature is zero degree celsius.

Substance	Mass	S.H.C./S.L.H.	Initial Temp.	Final Temp. = 0°C
Ice	428 g	?(L)	0°C	∴ θR = 0°С
Metal ball	400 g	0.6 Jg−1°C−1	600°C	∴ θF = (600 − 0) = 600°С

Heat gained by ice to melt = mL_ice

= 428 g × L

Heat lost by the metal ball = mcθ_F

= 400 × 0.6 × 600

= 144000 J

Heat gained = Heat lost

428 g × L = 144000 J

\[L=\frac{144000}{428} \mathrm{Jg}^{-1}\]

∴ L = 336.4 Jg⁻¹