Question

When 150 g of ice at 0°C is mixed with 300 g of water
at 50°C. The resulting temperature of mixture is (ignore
heat loss to surrounding)
4.2°C
9.7°C
6.7°C
32.6°C​

Answer 1

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•Constants that we need:

→Heat of fusion of ice Hi = 333.55J/g

→Heat capacity of water is Cpw = 4.186J/g-K

→The final temperature of the water is Tf

→To melt the ice at 0C to water at 0C requires

ΔH1 = mHi = 150(333.55) = 50,032.5J

→To raise the temperature of this water from 0C to Tf requires

→ΔH2 = mCpwΔT = 150(4.186)(Tf) = 627.9Tf J

→So the total amount of heat energy required is

• ΔHt = ΔH1 + ΔH2

• ΔHt = (50,032.5 + 627.9Tf)J

→The heat released by the cooling of the 300g of water at 50C is:

• ΔH3 = mCpwΔT = 300(4.186)(50 – Tf) = 62,790 - 1255.8Tf

→Then 1883.7Tf = 12,757.5 so Tf = 6.77

The final temperature of the mixture is 6.77C

Answer 2

Given:

Temperature of 150 gm of ice = 0°C

Temperature of 300 gm of water = 50°C

To find:

The resulting temperature of the mixture.

Solution:

Heat lost by water = Heat gained by ice to melt+ heat gained by the ice after melting

The specific heat of water = 1 cal/g°C

And the latent heat of fusion is 80 cal/g

mcΔT = m'L + m'cΔT

300*1*(50-T)= 150*80+ 150*1*(T-0)

On solving we get T = 6.7°C

Therefore the resulting temperature of the mixture is 6.7°C, thus the correct option is C.