Heat required to convert one gram of ice at 0°C into steam at 100°C is ( given L_{steam} = 5

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Heat required to convert one gram of ice at $0°C$ into steam at $100°C$ is $($given $L_{steam}$ $= 536\, cal/gm)$

A

$100$ calorie

B

$0.01$ kilocalorie

C

$716$ calorie

D

$1$ kilocalorie

Solution

(c) Conversion of ice $(0°C)$ into steam $(100°C)$ is as follows

Heat required in the given process $ = {Q_1} + {Q_2} + {Q_3}$

$ = 1 \times 80 + 1 \times 1 \times (100 – 0) + 1 \times 536 = 716\,cal$

Standard 11

Physics

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