Question

If you pour 0.600 kg of 20.0ºC water onto a 1.20-kg block of ice (which is initially at 0 ºC), what is the final temperature? You may assume that the water cools so rapidly that effects of the surroundings are negligible. The specific heat of water is 4 186 J/(kg * ºC), the specific heat of ice is 2 090 J/(kg * ºC), and the heat of fusion of water is 334 x10³ J/kg.

Answer 1

So if effect of surrounding is considered to be negligible then there is not heat exchange between surrounding and water ice system so we can write,

so mi mass of ice

T1= 0°C= 273K

mw is mass of water

Cw is specific heat of water

T2= 20°C= 20+273= 293K

L is latent heat of fusion

Now because no heat exchange occur with surrounding so delta Q=0

So, plug in values we get Tf

So we calculate

We get

Tf =12.6°C final Temprature.