Heat required to convert one gram of ice at $0^{_{-}^{0}} C$ into steam at $100^{_{-}^{0}} C$ is (given $L_{s t e a m}$ = 536 cal/gm)-

100 calorie

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0.01 kilocalorie

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716 calorie

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1 kilocaorie

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Solution

The correct option is C 716 calorie
It is known that $Q = m . c Δ θ$
Conversion of ice $(0^{\circ} C)$ into steam $(100^{\circ} C)$ is as follows:
Heat required in the given process $= Q_{1} + Q_{2} + Q_{3}$
putting values
$Q_{t o t a l} = 1 \times 80 + 1 \times 1 \times (100 - 0) + 1 \times 536 = 716 c a l$

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Heat required to convert one gram of ice at $0^{\circ} C$ into steam at $100^{\circ} C$ is (given $L_{s t r e a m}$ = 536 cal/gm)

^{0} C

^{0} C

10^{0} C

100^{0} C

536 c a l / g m - K

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