100 g of ice latent heat 80 cal g-1 at 0-C is mixed with 100 g of water specific heat 1 cal g-1 -C-1 at 80-C- The final temperature of the mixture will be-

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Calorimetry

100 g of ice ...

Question

$100 g$ of ice $(latent heat 80 {cal g}^{- 1})$ at $0^{\circ} C$ is mixed with $100 g$ of water (specific heat $1 {cal g}^{- 1 \circ} C^{- 1}$ ) at $80^{\circ} C$ . The final temperature of the mixture will be:

0^{\circ} C

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40^{\circ} C

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80^{\circ} C

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< 0^{\circ} C

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Solution

The correct option is A $0^{\circ} C$
Given that,
Latent heat of fusion of ice $(L_{f}) = 80 cal/g$
Specific heat capacity of water $(c_{w}) = 1 {cal/g}^{\circ} C$
Mass of ice $(m_{i}) = 100 g$
Mass of water $(m_{w}) = 100 g$
Initial temperature of ice $(T_{1}) = 0^{\circ} C$
Initial temperature of water $(T_{2}) = 80^{\circ} C$
Let the temperature of mixture be $T$ .

Heat released in melting $100 g$ of ice to water is: $Q_{1} = m_{i} L_{f} = 100 \times 80 = 8000 cal$
& Heat required to reduce the temperature of water from $80^{\circ} C$ to $0^{\circ} C$ is:
$Q_{2} = m_{w} c_{w} Δ T = 100 \times 1 \times (80 - 0) = 8000 cal$
$∵ Q_{1} = Q_{2}$
From the priciple of calorimetry, we can say that heat released from the cooling of water at $80^{\circ} C$ will be used to just melt the given mass of ice. At equilibrium, temperature of the system will be $0^{\circ} C$ .
Thus, option (a) is the correct answer.

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100 g

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Calorimetry

Standard XII Physics

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100 g of ice (latent heat 80 cal g−1) at 0∘C is mixed with 100 g of water (specific heat 1 cal g−1 ∘C−1) at 80∘C. The final temperature of the mixture will be:

$100 g$ of ice $(latent heat 80 {cal g}^{- 1})$ at $0^{\circ} C$ is mixed with $100 g$ of water (specific heat $1 {cal g}^{- 1 \circ} C^{- 1}$ ) at $80^{\circ} C$ . The final temperature of the mixture will be: