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Heat of Reaction

1.5 kg of wat...

Question

1.5 kg of water is in saturated liquid state at 2 bar

$(v_{f} = 0.001061 m^{3} / k g, u_{f} = 504.0, k J / k g, h_{f} = 505 k J / k g$ .

Heat is added in a constant pressure process till the temperature of water reaches $400^{o} C$

$(v = 1.5493 m^{3} / k g, u = 2967.0 k J / k g, h - 3277.0 k J / k g)$ .

The heat added (in $k J$ ) in the process is
4158

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Solution

The correct option is A 4158
$Method I:$

From the first law of thermodynamics
$δ q = d u + p d v$

$δ q = (u_{2} - u_{1}) + p (v_{2} - v_{1})$

$δ q = (2967 - 504) + 2 \times 10^{2}$

$1.5493 - 0.001061$

$δ q = 2463 + 309.6478 = 2772.6478 k J / k g$

$For 1.5 k g of water,$

$Q = 2772.6478 \times 1.5 = 4158.97 k J$

$Method II:$

As we know, heat transfer in a constant pressure system is equal to change in enthalpy.

$(δ Q) = m (h_{2} - h_{1}) = 1.5 (3277 - 505) = 4158 k J$

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