Question

Water of volume 2 L in a container is heated with a coil of 1 kW at 27$^{\circ}$C. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27$^{\circ}$C to 77$^{\circ}$C? [Specific heat of water is 4.2 kJ/kg]

• A. 8 min 20 s
• B. 6 min 2 s
• C. 7 min
• D. 14 min

Answer 1

Answer: (A)

8 min 20 s

Answer 2

Energy gained by water (in 1 s)
= energy supplied - energy lost = 1000 J - 160 J = 840 J
Total heat required to raise the temperature of water from
27$^{\circ}$C to 77$^{\circ}$C is ms$\Delta \theta$
Hence, the required time
$t=\frac{ms \Delta \theta}{rate \, by \, which \, energy \, is \, gained \, by \, water}$
$\frac{(2)(4.2 \times 10^3)(50)}{840}=500 s =8 min 20 s$
$\therefore \,$ Correct answer is (a)