Question

In preparing dinner you need a cup of very hot water, which you prepare on your electric stove.  You use 40 kJ of electrical energy to heat 300 g of water starting at 20 °C.  What is the final temperature of the water?

Answer 1

Heat(Q)=40kJ

Mass(m)=300g

Specific heat capacity of water(c)=4.18kJ/Kg

1000g=1kg

300g =?

$\frac {300g×1kg}{1000g}=0.3kg$

But Q=mc$\theta$

40kJ=0.3kg×4.18kJ/kg×(x-20$\degree$ C)

40 kJ=1.254x-25.08kJ/kg $\degree$ C

$\frac{1.254x}{1.254}=\frac{65.08}{1.254}$

x=51.9$\degree$ C