Question

Calculate the temperature of a furnace, when a 400 g of a copper ball, taken out from it, melts only 400 g of ice to form water at 0°C. Take S.H.C. of copper = 0.4 Jg⁻¹°C⁻¹ and S.L.H. of ice = 336 Jg⁻¹.

Answer 1

As the ice melts to form water at 0°C, therefore, final temperature is zero degree celsius.

Substance	Mass	S.H.C./S.L.H.	Initial Temp.	Final Temp. = 0°C
Ice	400 g	336 Jg−1	0°C	∴ θR = 0°С
Copper ball	400 g	0.4 Jg−1°C−1	? (θ)	∴ θF = (θ − 0) = θ

Heat gained by ice to melt = mL_ice

= 400 × 336

= 134400 J

Heat lost by the copper ball = mcθ_F

= 400 × 0.4 × θ

= 160 θ

Heat gained = Heat lost

134400 = 160 θ

\[\theta = \frac{134400}{160} \]

∴ θ = 840°C