Question

400 g of mercury of specific heat capacity 0.14 Jg⁻¹°C⁻¹ is heated by a 200 W heater for 1 min. and 40 s. If initially mercury is at 0°C, calculate its final temperature.

Answer 1

Time, t = 1 min and 40 s = 100 s

Energy supplied by the heater (E) = P × t

= 200 × 100

= 20000 J

Heat gained by mercury Q = mc(θ_f − θ_i)

Since θ_i = 0°C, Q = mcθ_f

$$\theta_f=\frac{Q}{mc}$$

$$=\frac{20000}{400 \times 0.14}$$

$$=\frac{20000}{56}$$

= 357.1°C