Gujarati
Hindi
10-1.Thermometry, Thermal Expansion and Calorimetry
medium

Heat needed to convert $1$ kilogram ice at $0^o\,C$ to steam at $100^o\,C$ is

A

$1\ kcal$

B

$720\ kcal$

C

$1\ cal$

D

$720\ cal$

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 $100^{\circ} \mathrm{C}$

$=80+100+540=720 \mathrm{kcal}$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.