Question

40 g of hot water is poured in 100 g of cold water, when temperature of cold water rises by $10{}^{o}C$. If the temperature of hot water is $60{}^{o}C$, calculate the initial temperature of cold water.

Answer 1

For Hot water:
$Mass=40g$
$S.H.C=4.2J{g}^{-1}{}^o{C}^{-1}$
$InitialTemp.=60{}^{o}C$
$FinalTemp.=x$

$△T_H=(60-x)$
for Cold water:
$Mass=100g$
$S.H.C=4.2J{g}^{-1}{}^o{C}^{-1}$
$InitialTemp.=?$
$FinalTemp.=x$

$△T_C={10}^{o}C$
$Heatgivenoutbyhotwater=mc△T_H$
$=40\times 4.2\times (60-x)$
$Heatabsorbedbycoldwater=mc△T_C=100\times 4.2\times 10$
$\therefore Heatgivenoutbyhotwater=Heatabsorbedbycoldwater$
$\therefore 40\times 4.2\times (60-x)=100\times 4.2\times 10$
$2400-40x=1000$
$40x=2400-1000$
$\therefore x=\frac{1400}{40}=35{}^{o}C$
$Thus,finaltemperatureofmixture=35{}^{o}C$
$Initialtemperatureofcoldwater=(35-10)=25{}^{o}C$

$Sothecorrectansweris25$