Question

If $400g$ of water at ${40}^0$ temperature is mixed with $100g$ of water at ${30}^0$ of temperature, Calculate the resulting temperature of the water after mixing?

• A. ${13}^o$
• B. ${26}^o$
• C. ${36}^o$
• D. ${38}^o$
• E. ${44}^o$

Answer 1

The correct option is D ${38}^o$

Given : $m_1=400$ g $T_1={40}^{o}C$ $m_2=100$ g $T_2={30}^{o}C$

Heat released from 400 g of water at ${40}^{o}C$ is completely absorbed by the 100 g of water at ${30}^{o}C$ such that net heat exchanged between system and surrounding is zero.

Let the specific heat and equilibrium temperature of water be $S$ and $T^{\prime }$ respectively.

$\therefore$ $m_{1}S(T^{\prime }-T_1)+m_{2}S(T^{\prime }-T_2)=0$

OR $400\times S(T^{\prime }-40)+100\times S(T^{\prime }-30)=0$

OR $5T^{\prime }=190$ $⟹T^{\prime }={38}^{o}C$