A) 1 gm
B) 1 kg
C) 10 gm
D) 10 kg
Correct Answer: B
Solution :
Suppose m kg steam required per hour Heat released by steam in following three steps (i) When 150°C steam \[\xrightarrow[{{Q}_{1}}]{}100{}^\circ C\] steam Q1 = mcSteam Dq = m ´ 1 (150 ? 100) = 50 m cal (ii) When 150°C steam \[\xrightarrow[{{Q}_{2}}]{}100{}^\circ C\] water Q2 = mLV = m ´ 540 = 540 m cal (iii) When 100°C water \[\xrightarrow[{{Q}_{2}}]{}90{}^\circ C\] water Q3 = mcW Dq = m ´ 1 ´ (100 ? 90) = 10 m cal Hence total heat given by the steam Q = Q1 +Q2 + Q3 = 600 mcal ... (i) Heat taken by 10 kg water \[Q'=m{{c}_{W}}\Delta \theta =10\times {{10}^{3}}\times 1\times (80-20)=600\times {{10}^{3}}cal\] Hence Q = Q¢ Þ 600 m = 600 ´ 103 Þ m = 103 gm = 1kg.You need to login to perform this action.
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