Question

$19\, g$ of water at $30? C$ and $5\, g$ of ice at $- 20?\, C$ are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice = $0.5\,cal g^{-1} ?C ^{-1}$ and latent heat of fusion of ice = $80\, cal \,g^{-1}$

• A. 0? C
• B. - 5? C
• C. 5? C
• D. 10? C

Answer 1

Answer: (C)

5? C

Answer 2

Let the final temperature of mixture be $t^{\circ} C$.
Heat lost by water in calories,
$H_{1} =19 \times 1 \times(30-t)$
$=570-19 t$
Heat taken by ice, $H _{2}=m s_{ p } \Delta t+m L+m s_{w} t$
$=5 \times(0.5) 20+5 \times 80+5 \times 1 \times t$
According to principle of calorimetry,
$H_{1}=H_{2}$
$5 \times(0.5) \times 20+5 \times 80+5 t=570-19 t$
$\Rightarrow 24 t=570-450=120$
$\Rightarrow t=5^{\circ} C$