ICSE Class 10 Physics 2014 Paper

Question : 28 of 66

Marks: +1, -0

50\ \text{g}

of metal piece at

27^{\circ}\ \text{C}

requires

2400\ \text{J}

of heat energy so as to attain a temperature of

327^{\circ}\ \text{C}

. Calculate the specific heat capacity of the metal.

Solution:

Given

m=50\ \text{g} ; \Delta t=327^{\circ}\ \text{C}-27^{\circ}\ \text{C}=300^{\circ}\ \text{C}\;\text{, }\;

Q=2400\ \text{J}.

\;\text{ Now, }\;\; Q=m c \Delta t

\;2400=50 \times c \times 300

\;c=\;\frac{2400}{50 \times 300}=0.16\ \text{J} / \text{g}\,{}^{\circ}\text{C}.

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