A) 18.2ºC
B) 22ºC
C) 20.2ºC
D) 25.2ºC
Correct Answer: C
Solution :
Heat gain = heat lost CA(16 ?12) = CB (19 ? 16) Þ \[\frac{{{C}_{A}}}{{{C}_{B}}}\] = \[\frac{3}{4}\] and CB(23 ? 19) = CC (28 ? 23) Þ\[\frac{{{C}_{B}}}{{{C}_{C}}}\] = \[\frac{5}{4}\] Þ \[\frac{{{C}_{A}}}{{{C}_{C}}}=\frac{15}{16}\] ...(i) If q is the temperature when A and C are mixed then, \[{{C}_{A}}(\theta -12)\] = \[{{C}_{C}}(28-\theta )\] Þ \[\frac{{{C}_{A}}}{{{C}_{C}}}\]\[=\frac{28-\theta }{\theta -12}\] ...(ii) On solving equation (i) and (ii) q = 20.2ºC.
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