A cannot engine extracts heat from water at 00C and rejects it to room at 24.40C. The work required by the refrigerator for every 1 kg of water converted into ice (latent heat of ice = 336kj/kg ) is

T2=273k,T1=297.4kw=?m−1kg;Lice=336000J/kgQ2w=T2T1−T2⇒mLice W=T2T1−T21×336000w=27324.4w=336000×24.4273=30.03×103J=30kJ

A cannot engine extracts heat from water at 00C and rejects it to room at 24.40C. The work required by the refrigerator for every 1 kg of water converted into ice (latent heat of ice = 336kj/kg ) is

T2=273k,T1=297.4kw=?m−1kg;Lice=336000J/kgQ2w=T2T1−T2⇒mLice W=T2T1−T21×336000w=27324.4w=336000×24.4273=30.03×103J=30kJ

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A cannot engine extracts heat from water at 0⁰C and rejects it to room at 24.4⁰C. The work required by the refrigerator for every 1 kg of water converted into ice (latent heat of ice = 336kj/kg ) is

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336 KJ

24.4KJ

11.2KJ

30 KJ

answer is A.

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Detailed Solution

$\begin{array}{l} T_{2} = 273 k, T_{1} = 297 .4 k \\ w = ? m - 1 kg; L_{ice} = 336000 J / kg \\ \frac{Q_{2}}{w} = \frac{T_{2}}{T_{1} - T_{2}} \Rightarrow \frac{{mL}_{ice}}{W} = \frac{T_{2}}{T_{1} - T_{2}} \\ \frac{1 \times 336000}{w} = \frac{273}{24 .4} \\ w = \frac{336000 \times 24 .4}{273} \\ = 30 .03 \times 10^{3} J = 30 kJ \end{array}$