Question

A metal ball of 0.20 kg and at 200°C, when placed on an ice block melts 100 g of ice, when its temperature stops falling. If sp. latent heat of ice is 340 Jg⁻¹, calculate specific heat capacity of the metal ball.

Answer 1

As the temperature of the metal ball stops falling, therefore, final temperature is zero degree celsius.

Substance	Mass	S.H.C./S.L.H.	Initial Temp.	Final Temp. = 0°C
Ice	100 g	340 Jg−1	0°C	∴ θR = 0°С
Metal ball	0.20 kg = 200 g	? (c)	200°C	∴ θF = (200 − 0) = 200°C

Heat gained by ice to melt = mL_ice

= 100 × 340

= 34000 J

Heat lost by the metal ball = mcθ_F

= 200 × c × 200

= 40000 c

Heat gained = Heat lost

40000 c = 34000

\[c = \frac{34000}{40000} \ \mathrm{Jg^{-1}{}^{\circ}C^{-1}} \]

∴ c = 0.85 Jg⁻¹°C⁻¹

Mass of the metal ball = 0.20 kg = 200 g

Heat gained by ice = 100 × 340 = 34000 J

If c is the specific heat capacity, heat lost by the ball = 200 × c × (200 − 0) = 40000 c

Equating, 40000 c = 34000

c = \[\frac{34000}{40000}\]

c = 0.85 Jg⁻¹°C⁻¹