A steel billet of 2000 kg mass is to be cooled from 250 k to 450 K. The heat released during this process is to be used as a source of energy. The ambient temperature is 303 K and specific heat of steel is 0.5 kJ/kgK. The available energy of this billet is

30.95 MJ

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0.10 MJ

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10.35 MJ

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490.44 MJ

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Solution

The correct option is D 490.44 MJ
(a)
Given: Ambient temperature, $T_{0} = 303 K$
$Q = m c_{p} Δ T$
$= 2000 \times (0.5 k J / k g K) \times (1250 - 450)$
$= 800000 k J$
$Δ S = m c_{p} I n \frac{T_{1}}{T_{2}} = 2000 \times 0.5 \times I n \frac{1250}{450}$
$= 1021.165 k J / K$
Available energy for the billet
$A . E . = Q - T_{0} Δ A = 800000 - 303 \times 1021.165$
$= 490587.005 k J = 490.58 M J$

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