Question

A metal ball of mass 0.5 kg and at 900°C is placed on a block of ice, till it attains the temperature of ice. If the S.H.C. of the metal ball is 850 Jkg⁻¹K⁻¹, calculate the amount of ice, which melts. Take S.L.H. of ice = 34 × 10⁴ Jkg⁻¹.

Answer 1

As the metal ball attains the temperature of ice, therefore, final temperature is zero degree celsius.

Substance	Mass	S.H.C./S.L.H.	Initial Temp.	Final Temp. = 0°C
Ice	? (m)	34 × 104 Jkg−1	0°C	∴ θR = 0°С
Metal ball	0.5 kg	850 Jkg−1K−1	900°C	∴ θF = (900 − 0) = 900°С

Heat gained by ice to melt = mL_ice

= m × 34 × 10⁴

Heat lost by the metal ball = mcθ_F

= 0.5 × 850 × 900

= 382500 J

Heat gained = Heat lost

m × 34 × 10⁴ = 382500

\[m = \frac{382500}{340000} \ \text{kg} \]

∴ m = 1.125 kg