Question

$2$ kg water at ${100}^{\circ }C$ and $2.5$ kg water at ${50}^{\circ }C$ is kept in two identical containers $A$and $B$ respectively of water equivalent $0.5$ kg. If water of container $A$ is placed poured into container $B$ the final temperature of mixture is $T_1$ and if water of container $B$ is poured into container $A$ the final temperature is $T_2$ (heat loss is negligible)

• A. $T_1=T_2$
• B. $T_2-T_1=5$
• C. $T_2={75}^{\circ }C$
• D. $T_1={75}^{\circ }$

Answer 1

Answer: (B), (C)

The correct options are
B $T_2-T_1=5$
C $T_2={75}^{\circ }C$

Water equivalent of a body is the mass of water, which would absorb or evolve the same amount of heat as is done by the body in rising or falling through the same range of temperature.

It means, when water from A is poured in B, container B has an extra amount of water, equal to the water equivalent of container B because temperature of B rises through the same range as of the water in it, the same happens when water is poured from B to A.

Now, by the principle of mixtures, heat lost = heat gained

When water is poured from A to B:

$\Rightarrow m_{A}c\Delta t_A=(m_B+0.5)c\Delta t_B$

$\Rightarrow 2c(100-T_1)=(2.5+0.5)c(T_1-50)$

$\Rightarrow 200-2T_1=3T_1-150$

$\Rightarrow 5T_1=350$

$\Rightarrow T_1=350/5={70}^{o}C$

When water is poured from B to A:

$\Rightarrow (m_A+0.5)c\Delta t_A=m_{B}c\Delta t_B$

$\Rightarrow (2+0.5)c(100-T_2)=2.5c(T_2-50)$

$\Rightarrow 100-T_2=T_2-50$

$\Rightarrow 2T_2=150$

$\Rightarrow T_2=150/2={75}^{o}C$

Hence , $T_2-T_1=75-70=5^{o}C$