Calculate the quantity of heat required to convert 10 kg of ice at 0- C into water at the same temperature- Latent heat of fusion of ice -335 - 103 J - kg A- 33-5 kJB- 33-5 kJC- 3350 kJD- 3-35 kJ

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Standard X

Physics

Specific Heat Capacity

Calculate the...

Question

Calculate the quantity of heat required to convert 10 kg of ice at $0^{\circ} C$ into water at the same temperature.
(Latent heat of fusion of ice $- 335 \times 10^{3} J / k g$ )

33.5 kJ

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33.5 kJ

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3350 kJ

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3.35 kJ

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Solution

The correct option is C 3350 kJ
We know, $Q = m L_{f}$
(where Q - quantity of heat required, m - mass, $L_{f}$ - latent heat of fusion.)
Given m = 10 kg
$L_{f} = 335 \times 10^{3} J / k g$
Quantity of heat required $= m L_{f}$
$= 10 \times 335 \times 10^{3}$
$= 3350 \times 10^{3}$
$= 3350 k J$

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Calculate the quantity of heat required to convert 10 kg of ice at 0∘C into water at the same temperature. (Latent heat of fusion of ice −335×103J/kg)

The correct option is C 3350 kJWe know, Q=mLf (where Q - quantity of heat required, m - mass, Lf - latent heat of fusion.) Given m = 10 kg Lf=335×103 J/kg Quantity of heat required =mLf =10×335×103 =3350×103 =3350 kJ

Calculate the quantity of heat required to convert 10 kg of ice at $0^{\circ} C$ into water at the same temperature.
(Latent heat of fusion of ice $- 335 \times 10^{3} J / k g$ )

The correct option is C 3350 kJ
We know, $Q = m L_{f}$
(where Q - quantity of heat required, m - mass, $L_{f}$ - latent heat of fusion.)
Given m = 10 kg
$L_{f} = 335 \times 10^{3} J / k g$
Quantity of heat required $= m L_{f}$
$= 10 \times 335 \times 10^{3}$
$= 3350 \times 10^{3}$
$= 3350 k J$