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