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Latent Heat

Determine the...

Question

Determine the heat required to convert $3 k g$ of ice from $- 7^{\circ} C$ to $40^{\circ} C$ . Latent heat of ice = $334.7 J / k g K$ , specific heat of ice = $2108 J / k g K$ and specific heat of water = $4186 J / k g K$ .

448 k J

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548 k J

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934 k J

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487 k J

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Solution

The correct option is B $548 k J$
Given:
Mass of ice, $m = 3 k g$
Initial temperature, $T_{i} = - 7^{\circ} C$
Final temperature, $T_{f} = 40^{\circ} C$
Latent heat of ice, $L = 334.7 J / k g K$
Specific heat of ice, $S_{i} = 2108 J / k g K$
Specific heat of water, $S_{w} = 4186 J / k g K$
Heat required to raise the temperature of ice from $- 7^{\circ} C$ to $0^{\circ} C$ , $Q_{1} = m \times S_{i} \times [0 - (T_{i})]$
$\Rightarrow 3 \times 2108 \times [0 - (- 7)] = 44268 J$
Heat required to melt 3 kg of ice, $Q_{2} = m \times L$
$\Rightarrow 3 \times 334.7 = 1004 J$
Heat required to raise the temperature of water from $0^{\circ} C$ to $40^{\circ} C$ , $Q_{3} = m \times S_{w} \times (40 - 0)$
$\Rightarrow 3 \times 4186 \times 40 = 502, 320 J$
Total heat required = $Q_{1} + Q_{2} + Q_{3}$
$\Rightarrow 44268 + 1004 + 502320 = 547592 J$
We know that $1000 J = 1 k J$
$∴ 547592 J = 547.592 k J$
which is $\approx 548 k J$

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Determine the heat required to convert 3 kg of ice from −7∘C to 40∘C. Latent heat of ice = 334.7 J/kg K, specific heat of ice = 2108 J/kg K and specific heat of water = 4186 J/kg K.

Determine the heat required to convert $3 k g$ of ice from $- 7^{\circ} C$ to $40^{\circ} C$ . Latent heat of ice = $334.7 J / k g K$ , specific heat of ice = $2108 J / k g K$ and specific heat of water = $4186 J / k g K$ .