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Calorimeter

When 1.5 kg o...

Question

When 1.5 kg of ice at $0$ mixed with 2 kg of water at $70$ in a container, the resulting temperature is $5$ .Then the heat of fusion of ice is $(s_{w a t e r} = 4186 J {k g}^{- 1} K^{- 1})$

1.42 \times 10^{5} J {k g}^{- 1}

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2.42 \times 10^{5} J {k g}^{- 1}

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3.42 \times 10^{5} J {k g}^{- 1}

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4.42 \times 10^{5} J {k g}^{- 1}

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Solution

The correct option is B $3.42 \times 10^{5} J {k g}^{- 1}$
Heat lost by water $= m_{w} s_{w} (T_{i} - T_{f})$

$= 2 \times 4186 \times (70 - 5) = 544180 J$

Heat required to melt ice $= m_{f} L_{f} = 1.5 \times L_{f}$

The heat required to raise the temperature of the ice

$= m_{i} s_{w} (T_{f} - T_{0}) = 1.5 \times (4186) \times (5 - 0) = 31395 J$

By the principle of calorimetry

$H e a t l o s t = h e a t g a i n e d$

$544180 = 1.5 L_{f} + 31395$

$∴ L_{f} = \frac{512785}{1.5}$

$= 341856.67 = 3.42 \times 10^{5} J {k g}^{- 1}$

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