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Q.

Take 100 g of hot water at $80°C$ in thermosflask. Add 200 g of cold water at $20°C$ in the thermoflask. Stir the contents and measure the temperature of the mixture. Then, the resultant temperature is [No heat lost or gained]

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a

40°C

b

20°C

c

80°C

d

60°C

answer is A.

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Detailed Solution

m_{h} = 100 g

of water

$t_{1} = 80 ° C; m_{c} = 200 g; t_{2} = 20 ° C$

We know $T_{R} = \frac{m_{1} T_{1} + m_{2} T_{2}}{m_{1} + m_{2}}$ (when two materials are same)

$T_{R} = \frac{100 \times 80 + 200 \times 20}{100 + 200} = \frac{120}{3} = 40 ° C$

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Take 100 g of hot water at80°C in thermosflask. Add 200 g of cold water at20°C in the thermoflask. Stir the contents and measure the temperature of the mixture. Then, the resultant temperature is [No heat lost or gained]