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Area and Volume Expansion

Two liquids A...

Question

Two liquids A and B are at temperatures $75^{o} C$ and $150^{o} C$ , respectively. Their masses are in the ratio 2 : 3 and specific heats are in the ratio 3 : 4. The resultant temperature of the mixture, when the above liquids are mixed (neglect the water equivalent of container ) is :

75^{o} C

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100^{o} C

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50^{o} C

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125^{o} C

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Solution

The correct option is D $125^{o} C$
Heat lost by one liquid $=$ Heat gained by another liquid
$m_{1} . C_{1} . Δ T_{1} = m_{2} . C_{2} . Δ T_{2}$

$m_{1} . C_{1} (T - 75) = m_{2} . C_{2} (150 - T)$

$\frac{m_{1}}{m_{2}} . \frac{c_{1}}{c_{2}} = \frac{150 - T}{T - 75}$

$\frac{2}{3} . \frac{3}{4} = \frac{150 - T}{T - 75}$

$\frac{1}{2} = \frac{150 - T}{T - 75}$

$T - 75 = 300 - 2 T$

$3 T = 375$

$T = 125^{o} C$

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Two liquids A and B are at temperatures 75oC and 150oC, respectively. Their masses are in the ratio 2 : 3 and specific heats are in the ratio 3 : 4. The resultant temperature of the mixture, when the above liquids are mixed (neglect the water equivalent of container ) is :

The correct option is D 125oCHeat lost by one liquid = Heat gained by another liquidm1.C1.ΔT1=m2.C2.ΔT2m1.C1(T−75)=m2.C2(150−T)m1m2.c1c2=150−TT−7523.34=150−TT−7512=150−TT−75T−75=300−2T3T=375T=125oC

Two liquids A and B are at temperatures $75^{o} C$ and $150^{o} C$ , respectively. Their masses are in the ratio 2 : 3 and specific heats are in the ratio 3 : 4. The resultant temperature of the mixture, when the above liquids are mixed (neglect the water equivalent of container ) is :