Question

Find out the resulting temperature of the water, If $360g$ of water at ${95}^{\circ }C$ is mixed with $275g$ of water at ${10}^{\circ }C$ ?

• A. ${37}^{\circ }C$
• B. ${49}^{\circ }C$
• C. ${58}^{\circ }C$
• D. ${70}^{\circ }C$
• E. ${82}^{\circ }C$

Answer 1

The correct option is C ${58}^{\circ }C$

By principle of mixtures ,

Heat lost = Heat gained

$c{m}_2(t_2-t)=c{m}_1(t-t_1)$ , where $t$ is final temperature of mixture ,

now, given - $m_1=275g,m_2=360g,t_1={10}^{o}C,t_2={95}^{o}C$

therefore ,

$360(95-t)=275(t-10)$

or $t={58.18}^{o}C$