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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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An engine intakes 10 g of steam per sec at $\;100^{0}C\;$ & cools it down to $\;50^{0}C\;$. ( latent heat of vapourization of steam = 540 $\;calg^{-1}$) Heat rejected by engine per minute

$(a)\;5.4 \times 10^{3}\;cal\qquad(b)\;0.5 \times 10^{3}\;cal\qquad(c)\;4.9 \times 10^{3}\;cal\qquad(d)\;5.9 \times 10^{3}\;cal$

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Answer : (d) $\;5.9 \times 10^{3}\;cal$
Explanation :
Heat rejected during condensation of steam = $\;10 \times 540 = 5.4 \times 10^{3}\;cal$
Heat rejected during cooling = $\;10 \times 1 \times (100-50)$
$=0.5 \times 10^{3} \;cal$
Therefore , Total heat rejected per minute
$=5.9 \times 10^{3} \;cal$
answered Mar 12, 2014 by yamini.v
 

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