Question

Calculate the initial temperature of liquid $A$ of mass $400$g having a specific heat capacity of $2.4J{kg}^{-1}{K}^{-1}$ when it is mixed with liquid $B$ of mass $750$g having a specific heat capacity of $1.6J{kg}^{-1}{K}^{-1}$ at ${40}^{\circ }C$. The final temperature of the mixture becomes ${25}^{\circ }C$.

• A. ${15}^{\circ }C$
• B. ${7.5}^{\circ }C$
• C. ${4.2}^{\circ }C$
• D. ${6.25}^{\circ }C$

Answer 1

The correct option is D ${6.25}^{\circ }C$

Let the initial temperature of $A$ be $t^{\circ }C$
Rise in temperature of $A$ $={(25-t)}^{\circ }C$
Mass of $A=400g=0.4kg$
Specific heat capacity of A $=2.4J{kg}^{-1}{K}^{-1}$
Heat taken by A $=0.4\times 2.4\times (25-t)J$
Initial temperature of B $={40}^{\circ }C$
Fall in temperature of B $={(40-25)}^{\circ }C={15}^{\circ }C$Mass of B =750g = 0.75kg
Specific heat capacity of B $=1.6J{kg}^{-1}{K}^{-1}$
Heat given by B $=0.75\times 1.6\times 15J=18J$
Assuming no heat is lost,Heat taken by A = Heat given by B
$0.4\times 2.4\times (25-t)J=18J$
$24-0.96t=18$
$0.96t=24-18$
$t={6.25}^{\circ }C$