Question

Find the result of mixing 10 g of ice at -${10}^{\circ }C$ with 10 g of water at ${10}^{\circ }C$.Specific heat capacity of ice=2.1 J $g^{-1}{K}^{-1}$,specific latent heat of ice=336 J $g^{-1}$ and specifc heat capacity of water =4.2 J $g^{-1}{K}^{-1}$.

Answer 1

Let the whole ice melt and let the final temperature of the mixture be T^oC.
Amount of heat energy gained by 10 g of ice at -10^oC to raise its temperature to 0^oC= 10 $\times$ 10 $\times$ 2.1 = 210 J
Amount of heat energy gained by 10 g of ice at 0^oC to convert into water at 0^oC = 10 $\times$ 336 = 3360 J
Amount of heat energy gained by 10 g of water (obtained from ice) at 0^oC to raise its temperature to T^oC = 10 $\times$ 4.2 $\times$ (T-0)=42T
Amount of heat energy released by 10 g of water at 10^oC to lower its temperature to T^oC = 10 $\times$ 4.2 $\times$ (10-T) = 420-42T
Heat energy gained = Heat energy lost
210 + 3360 + 42T = 420-42T
T = -37.5^oC
This cannot be true because the equilibrium temperature can't be equal to this.
So, complete ice does not melt.
Let 'm' g of ice melt. The final temperature of the mixture becomes 0^oC.
So, amount of heat energy gained by 10g of ice at -10^oC to raise its temperature to 0^oC= 10 $\times$ 10 $\times$ 2.1 = 210 J
Amount of heat energy gained by m gm of ice at 0^oC to convert into water at 0^oC = m $\times$ 336 = 336m J
Amount of heat energy released by 10g of water at 10^oC to lower its temperature to 0^oC = 10 $\times$ 4.2 $\times$ (10-0)=420 J
Heat energy gained = Heat energy lost
210 + 336m = 420
m = 0.625 grams