A steam engine intakes 50 g of steam at 100$^\circ$C per minute and cools it down to 20$^\circ$C. If latent heat of vaporization of steam is 540 cal g$-$1, then the heat rejected by the steam engine per minute is __________ $\times$ 103 cal.
(Given : specific heat capacity of water : 1 cal g$-$1 $^\circ$C$-$1)
Answer (integer)
31
Solution
<p>$\Delta$Q<sub>rej</sub> = 50 $\times$ 540 + 50 $\times$ 1 $\times$ (100 $-$ 20)</p>
<p>= 50 $\times$ [540 + 80]</p>
<p>= 50 $\times$ 620</p>
<p>= 31000 cal</p>
<p>= 31 $\times$ 10<sup>3</sup> cal</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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