ICSE Class 10 Physics 2015 Solved Paper

Question : 67 of 74

Marks: +1, -0

A refrigerator converts

100 {\g}

of water at

20^{∘} {\C}

to ice at

-10^{∘} {\C}

in 35 minutes.
Calculate the average rate of heat extraction in terms of watts.
Given:
Specific heat capacity of ice

=2.1 {\J} {\g}^{-1}{ }^{∘} {\C}{ }^{-1}

.
Specific heat capacity of water

=4.2 {\J} {\g}^{-1}

{ }^{∘} {\C}^{-1}

.
Specific Latent heat of fusion of ice

=336 {\J}

g^{-1}

Solution:

\;\;\text " Mass of water "\;=100 {\g}=0.1 {\kg} .

\;\;\text " Temperature of water "\;=20^{∘} {\C} \;\text ". "\;

Amount of heat extracted to convert water from

20^{∘} {\C}

0^{∘} {\C}

{\Q}_1\;=m {\C}_{\;\text "water "\;} ∆ t

\;=0.1 × 4200 ×(20-0)

\;=8400 {\J}

Amount of heat extracted to convert water at

0^{∘} {\C}

0^{∘} {\C}

ice

Q_2\;=m L_{\;\text "ice "\;}=0.1 × 336 × 1000

\;=33600 {\J}

Amount of heat extracted to convert ice at

0^{∘} {\C}

to ice at

-10^{∘} {\C}

Q_3\;=m C_{\;\text "ice "\;} ∆ t

\;=0.1 × 2.1 × 10^3 ×(0-(-10))

\;=2100 {\J}

\;\text " Total Heat "\;(Q)\;=Q_1+Q_2+Q_3

\;=8400+33600+2100

\;=44100 {\J}

∴ \;\; {\P} t\;= {\Q}

\;\text " or "\; \;\; {\P}\;=\;{ {\Q}}/{t}

\;\text " Power "\;\;=\;{44100}/{35 × 60}=\;{44100}/{2100}

\;=21 {\Watt}

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