Heat is required to change 1 kg of ice at $- 20^{o}$ C into steam. $Q_{1}$ is the heat needed to warm the ice from $- 20^{o}$ C to $0^{o}$ C, $Q_{2}$ is the heat needed to melt the ice, $Q_{3}$ is the heat needed to warm the water from $0^{o}$ C to $100^{o}$ C and $Q_{4}$ is the heat needed to vaporize the water. Then

Q_{4} > Q_{3} > Q_{2} > Q_{1}

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Q_{4} > Q_{3} > Q_{1} > Q_{2}

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Q_{4} > Q_{2} > Q_{3} > Q_{1}

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Q_{4} > Q_{2} > Q_{1} > Q_{3}

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Solution

The correct option is A $Q_{4} > Q_{3} > Q_{2} > Q_{1}$
$Q_{1} = Heat needed to warm ice to get water$
$Q_{2} = Heat needed to melt the ice$
$Q_{3} = Heat needed to warm the water$
$Q_{4} = Heat needed to vaporize the water$
here $m = 1 k g$
$Vapour latent heat L_{v} = 540 c a l / g$
$Ice fusion latent heat L_{f} = 80 c a l / g$
$Specific heat of water s = 1 c a l / g^{0} C$
$Specific heat of ice s = 1 c a l / g^{0} C$
Thus, $Q_{1} = m \times s \times Δ T$
$Q_{1} = 1 \times 05 \times (0 - (- 20))$
$Q_{1} = 10 c a l$
$Q_{2} = m \times L_{f}$
$Q_{2} = 1 \times 80$
$Q_{2} = 80 c a l$
$Q_{3} = m \times s \times Δ T$
$Q_{3} = 1 \times 1 \times (100 - 0)$
$Q_{3} = 100 c a l$
$Q_{4} = m \times L_{v}$
$Q_{4} = 1 \times 540$
$Q_{4} = 540 c a l$
$Option A is correct option$

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