Question

A cube of ice of mass 55g at $0^{\circ }C$ is dropped in water weighing 220g at ${25}^{\circ }C$. what will the final temperatre of water be when the entice ice melts? (specific latent heat of ice = 80 cal/g, specific heat capacity of water = $1cal{g}^{{-1}^{\circ }}{C}^{-1}$)

• A. $4^{\circ }C$
• B. ${15}^{\circ }C$
• C. $8^{\circ }C$
• D. ${12}^{\circ }C$

Answer 1

Answer: (A)

$4^{\circ }C$

Given:
Heat energy absorbed by ice to melt = mL = 55 g x 80 cal/g = 4400 cal
Let the final temperature of water be $t^{\circ }C$
Heat energy required by ice to raise the temperature from $0^{\circ }C$ to $t^{\circ }C$
=mass x specific heat capacity x rise in temperature
$55g\times 1cal{g}^{{-1}^{\circ }}{C}^{-1}\times t^{\circ }C=55t^{\circ }C$.
Therefore, total energy absorbed by ice = (4400 + 55 t) cal
Heat is given by water and its temperature reduces from ${25}^{\circ }C$ to $t^{\circ }C$
=mass x specific heat capacity x fall in temperature
$=220g\times 1cal{g}^{1^{\circ }}{C}^{-1}\times {(25-t)}^{\circ }C=220(25-t)cal$
If there is no heat loss,
Heat given by water = heat taken by ice
220(25 - t) = (4400 + 55 t)
5500-220t = 4400 + 55 t
1100= 275 t
t =1100/275 = $4^{\circ }C$