JEE Main 2023 · Physics · Calorimetry and Heat Transfer · medium

50 g of ice at 0^circC is to be completely converted into water at 40^circC by passing steam at 100^circC into it in a perfectly insulated vessel of…

Numerical answer

50\ \mathrm{g}

of ice at

0^\circ\mathrm{C}

is to be completely converted into water at

40^\circ\mathrm{C}

by passing steam at

100^\circ\mathrm{C}

into it in a perfectly insulated vessel of negligible heat capacity. Taking latent heat of fusion

80\ \mathrm{cal\,g^{-1}}

, latent heat of vaporisation

540\ \mathrm{cal\,g^{-1}}

and specific heat of water

1\ \mathrm{cal\,g^{-1}\,{}^\circ C^{-1}}

, the mass of steam required (in g) is

Show answer and solution

Answer: 10

Heat absorbed by the ice: melting

50\times80=4000\ \mathrm{cal}

, then warming the water from

0

40^\circ\mathrm C

50\times1\times40=2000\ \mathrm{cal}

, total

6000\ \mathrm{cal}

. The steam of mass

m

gives up

540m

on condensing and

m\times1\times(100-40)=60m

on cooling to

40^\circ\mathrm C

, i.e.

600m

. Setting

600m=6000

gives

m=10\ \mathrm{g}

. (The final mixture is

60\ \mathrm g

of water at

40^\circ

C.)

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