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Temparature & Thermometer

Two solidbodi...

Question

Two solidbodies of equal mass $m$ initially at $T = 0^{\circ} C$ are heated at a uniform and same rate under identical conditions. The temperature of the first object with latent heat $L_{1}$ and specific heat capacity in solid state $C_{1}$ changes according to graph $1$ on the diagram. The temperature of the second object with latent heat $L_{2}$ and specific heat capacity in solid state $C_{2}$ changes according to graph $2$ on the diagram. Based on what is shown on the graph, the latent heats $L_{1} and L_{2}$ and the specific heat capacities $C_{1} and C_{2}$ in solid state obey which of the following relationships:

L_{1} < L_{2} : C_{1} > C_{2}

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L_{1} > L_{2} : C_{1} < C_{2}

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L_{1} > L_{2} : C_{1} > C_{2}

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L_{1} < L_{2} : C_{1} < C_{2}

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Solution

The correct option is B $L_{1} > L_{2} : C_{1} < C_{2}$
Formula used: $Q = m L, Q = m c Δ T$
If heat is supplied at constant rate $P,$ then
$Q = P Δ t$ and as during change of state
$Q = m L$
$∴ m L = P Δ t$
$i.e., L = [\frac{P}{m}] Δ t$
$Δ t$ is length of horizontal line in graph given.
Hence, $L_{1} > L_{2}$
$i.e. the ratio of latent heat of fusion of the two substance are in the ratio 3 : 4$ .

In the first slanted portion the substance is in solid state and its temparature is changing.
$Δ Q = m C Δ T$ and
$Δ Q = P Δ T$
So, $\frac{Δ T}{Δ t} = \frac{P}{m C}$
or, slope $= \frac{P}{m C}$
From graph slope of 1 is more than 2,
hence, $C_{1} < C_{2}$

FINAL ANSWER : (a).

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Temparature & Thermometer

Standard VI Chemistry

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