How much steam at 100- will just melt 64 gm of ice at -10-C-

We know that: nbsp;Q=mL, Q=msΔθ Given: m_1=64 g , θ_1= 10^∘C, s_1=0.5 cal/g^∘C, s_w=1 cal/g^∘C, nbsp; nbsp; θ_2=100^∘C, θ=0^∘C, L_v=540 cal/g, L_f=80 ca ...

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Standard VIII

Physics

Density

How much stea...

Question

How much steam at $100^{\circ}$ will just melt $64 g m$ of ice at $- 10^{\circ} C$ ?

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Solution

We know that: $Q = m L, Q = m s Δ θ$

Given: $m_{1} = 64 g, θ_{1} = - 10^{\circ} C, s_{1} = 0.5 c a l / g^{\circ} C, s_{w} = 1 c a l / g^{\circ} C, θ_{2} = 100^{\circ} C, θ = 0^{\circ} C, L_{v} = 540 c a l / g, L_{f} = 80 c a l / g$

Let $m_{2}$ be mass of steam required.
Heat lost by steam to reach $0^{\circ} C$ will be absorbed by ice during $- 10^{\circ} C$ to water at $0^{\circ} C$ ,

$m_{2} L_{v} + m_{2} s_{w} (100 - 0) = m_{1} \times s_{1} [0 - (- 10)] + m_{1} \times L_{f}$

$m_{2} \times 540 + m_{2} \times 1 \times 100 = 64 \times 0.5 \times 10 + 64 \times 80$

$640 m_{2} = 64 \times 85$

$m_{2} = 8.5 g$

Final answer: $8.5 g$

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Similar questions

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Density

Standard VIII Physics

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How much steam at 100∘ will just melt 64 gm of ice at −10∘C?

How much steam at $100^{\circ}$ will just melt $64 g m$ of ice at $- 10^{\circ} C$ ?