The graph shown in the figure represents change in the temperature of $5 k g$ of a substance with time, as it absorbs heat at a constant rate of $42 k J$ $m i n^{- 1}$ . The latent heat of vapourization of the substance is:

630 k J k g^{- 1}

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126 k J k g^{- 1}

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84 k J k g^{- 1}

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12.6 k J k g^{- 1}

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Solution

The correct option is C $84 k J k g^{- 1}$

$A B$ represents liquid to vapour phase change which occurs for $10 m i n$ and in each minute $42 k J$ of heat is supplied. Thus,
$△ Q = 42 \times 10 = 420 k J$

This heat vapourises $5 k g$ of the substance.
$5 L_{v} = 420 L_{v} = 84 k J k g^{- 1}$ .

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The graph shown in the figure represents change in the temperature of $5 k g$ of a substance with time, as it absorbs heat at a constant rate of $42 k J$ $m i n^{- 1}$ . The latent heat of vapourization of the substance is:

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The graph shown in the figure represents change in the temperature of 5 kg of a substance with time, as it absorbs heat at a constant rate of 42 kJ min−1. The latent heat of vapourization of the substance is:

The correct option is C 84 kJ kg−1 AB represents liquid to vapour phase change which occurs for 10 min and in each minute 42 kJ of heat is supplied. Thus, △Q=42×10=420 kJ This heat vapourises 5 kg of the substance. 5Lv=420Lv=84 kJ kg−1 .

The graph shown in the figure represents change in the temperature of $5 k g$ of a substance with time, as it absorbs heat at a constant rate of $42 k J$ $m i n^{- 1}$ . The latent heat of vapourization of the substance is:

The correct option is C $84 k J k g^{- 1}$

$A B$ represents liquid to vapour phase change which occurs for $10 m i n$ and in each minute $42 k J$ of heat is supplied. Thus,
$△ Q = 42 \times 10 = 420 k J$

This heat vapourises $5 k g$ of the substance.
$5 L_{v} = 420 L_{v} = 84 k J k g^{- 1}$ .