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  4. How much steam at 100° C will just melt 3200 g of ice at -10° C ? (Specific heat of ice =0.5 cal / g ° C, specific heat of water =1 cal / g ° C, latent heat of fusion of ice =80 cal / g, and latent heat of vaporisation of water =540 cal / g )

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Q. How much steam at $100^{\circ} C$ will just melt $3200 \,g$ of ice at $-10^{\circ} C ?$ (Specific heat of ice $=0.5 \,cal / g ^{\circ} C$, specific heat of water $=1 \,cal / g ^{\circ} C$, latent heat of fusion of ice $=80 \,cal / g$, and latent heat of vaporisation of water $=540 \,cal / g$ )

Thermal Properties of Matter

Solution:

Let $m$ grams be the mass of the steam.
Heat lost by the steam $=m \times L + m \times 1 \times(100-0)$
$=m \times 540+100 \,m=640 \,m$
Heat gained by ice $=m_{i} \times s \times \Delta T+m_{i} L$
$=3200 \times 0.5 \times[0-(-10)]+3200 \times 80 $
$=272000 \,cal$
According to the principle of calorimetry
$640\, m=272000$ or $ m=425 \,g$