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Heat needed to convert $1$ kilogram ice at $0^o\,C$ to steam at $100^o\,C$ is
$1\ kcal$
$720\ kcal$
$1\ cal$
$720\ cal$
Solution
Heat required to melt $1$ $kg$ ice into water at $0^{\circ} \mathrm{C}=80 \mathrm{kcal} .$ Heat required to raise temperature of $1\textrm{kg water from } 0 ^ { \circ } \mathrm { C } \text { to } 1 0 0 ^ { \circ } \mathrm { C }$
$=\mathrm{mS} . \Delta \mathrm{t}=(1 \mathrm{kg}) \times\left(1 \mathrm{kcal} / \mathrm{kg}^{\circ} \mathrm{C}\right) \times(100-0)$
$=100 \mathrm{kcal}$
Heat required to convert $1$ $kg$ boiling water into steam at $100^{\circ} \mathrm{C}$
$=\mathrm{mL}=1 \mathrm{kg} \times 540 \mathrm{kcal} / \mathrm{g}=540 \mathrm{kcal}$
Thus, total heat required to convert $1 \mathrm{kg}$ of ice at $0^{\circ} \mathrm{C}$ into steam at $100^{\circ} \mathrm{C}$
$=80+100+540=720 \mathrm{kcal}$