Q: Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of 0.1 g/sec . It melts completely in 100 sec . The rate of rise of temperature thereafter will be (Assume no loss of heat)
(a) 0.8 °C/sec
(b) 5.4 °C/sec
(c) 3.6 °C/sec
(d) None
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Ans: (a)
Sol: m/t = 0.1
m = 0.1 t = 0.1 × 100 = 10 gm
Q = mL
$\displaystyle \frac{dQ}{dt} = L\frac{dm}{dt}$
$\displaystyle \frac{dQ}{dt} = 80 \times 0.1 $
= 8 cal/sec
Again , $\displaystyle \frac{dQ}{dt} = m s \frac{d\theta}{dt} $
$\displaystyle \frac{d\theta}{dt} = \frac{\frac{dQ}{dt}}{m s} $
$\displaystyle \frac{d\theta}{dt} = \frac{8}{10 \times 1}$
= 0.8