M germs of steam at 100-C is mixed with 200 g of ice at its melting point in a thermally insulated container- If it produces liquid water at 40-C -heat of vaporization of water is 540 cal-g and heat of fusion of ice is 80 cal-g- the value of Min g is -

Using the principal of calorimetry, nbsp; Heat gained by the ice to become water at 40^∘C is equal to heat lost by the steam to nbsp;become water at 40^∘C ...

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Byju's Answer

Standard VII

Chemistry

Latent Heat of Vaporisation

M germs of st...

Question

M germs of steam at $100^{\circ} C$ is mixed with $200 g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} C$ [heat of vaporization of water is $540 cal/g$ and heat of fusion of ice is $80 cal/g$ ], the value of $M (in g)$ is .

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Solution

Using the principal of calorimetry,

Heat gained by the ice to become water at $40^{\circ} C$ is equal to heat lost by the steam to become water at $40^{\circ} C$

$m_{ice} L_{f} + m_{ice} (40 - 0) C_{w} = m_{stream} L_{v} + m_{stream} (100 - 40) C_{w}$

$(200 \times 80) + (200 \times 1 \times 40) = M (540) + M \times 1 \times (100 - 40)$

$\Rightarrow 600 M = 24000$

$\Rightarrow M = 40 g$

Hence, $40$ is the correct answer.

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M germs of steam at 100∘C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40∘C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M(in g) is .

M germs of steam at $100^{\circ} C$ is mixed with $200 g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} C$ [heat of vaporization of water is $540 cal/g$ and heat of fusion of ice is $80 cal/g$ ], the value of $M (in g)$ is .