Question

If you pour 0.0350 kg of 25.0°C water onto a 1.18 kg block of ice (which is initially at −15.0°C), what is the final temperature (in °C)? You may assume that the water cools so rapidly that effects of the surroundings are negligible.

Answer 1

Suppose the final temperature is T (Assuming final temperature is less than 0 C (T < 0) because amount of water is very low compared to amount of ice)

Now Using energy conservation:

Heat gained by Ice = Heat released by water

Q1 = Q2 + Q3 + Q4

Q1 = Heat absorbed by ice from -15 C to T C = mi*Ci*dT1

Q2 = Heat released by water from 25 C to 0 C = mw*Cw*dT2

Q3 = Heat released by water during phase change from water to ice = mw*Lf

Q4 = Heat released by water from 0 C to T C = mw*Cwi*dT3

mi*Ci*dT1 = mw*cw*dT2 + mw*Lf + mw*Ci*dT3

mi = mass of ice = 1.18 kg

mw = mass of water = 0.0350 kg

Lf = latent heat of fusion of ice = 3.34*10^5 J/kg

Cw = Specific heat capacity of water = 4186 J/kg-C

Ci = Specific heat capacity of ice = 2090 J/kg-C

dT1 = T - (-15) = T + 15

dT2 = 25 - 0 = 0

dT3 = 0 - T = -T

So Now using given values:

1.18*2090*(T + 15) = 0.0350*4186*25 + 0.0350*3.34*10^5 - 0.0350*2090*T

T = [1.18*2090*15 - 0.0350*4186*25 - 0.0350*3.34*10^5]/(-0.0350*2090 - 1.18*2090)

T = -8.52 C = final temperature of system

Let me know if you've any query.