Question

$1kg$ of water at ${20}^{o}C$ is, mixed with $800g$ of water at ${80}^{o}C$. Assuming that no heat is lost to the surroundings, calculate the final temperature of the mixture :

• A. ${24.44}^{o}C$
• B. ${46.67}^{o}C$
• C. ${44.44}^{o}C$
• D. ${54.44}^{o}C$

Answer 1

The correct option is B ${46.67}^{o}C$

Let the final temperature of the mixture is $t^{o}C$ .

Given $m_1=800g,m_2=1kg=1000g,t_1={80}^{o}C,t_2={20}^{o}C$

According to the principal of mixtures

Heat lost by water =Heat gained by water ,

$\Rightarrow$ $m_{1}c(t_1-t)=m_{2}c(t-t_2)$

$\Rightarrow$ $800(80-t)=1000(t-20)$

$\Rightarrow$ $4(80-t)=5(t-20)$

$\Rightarrow$ $320-4t=5t-100$

$\Rightarrow$ $9t=420$

$\Rightarrow$ $t=420/9={46.67}^{o}C$