Question

(a) Two 29 g ice cubes are dropped into 170 g of water in a thermally insulated container. If the water is initially at 26°C, and the ice comes directly from a freezer at -16°C, what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used? The specific heat of water is 4186 J/kg·K. The specific heat of ice is 2220 J/kg·K. The latent heat of fusion is 333 kJ/kg.

Answer 1

(a) First of all convert the masses in kilograms -
For the two ice cubes, mass = 2 * 0.029 = 0.058 kg
And for the water, mass = 0.170 kg

Let’s use the following equation to determine the amount of heat energy that is required to increase the temperature of the ice cubes from -16˚C to 0˚C.

Q = mass * specific heat * ∆ T
Q = 0.058 * 2220 * 16 = 2060.16 J
Let’s use equation to determine the amount of heat energy that is required to melt the ice.

Q = mass * heat of fusion
Q = 0.058 * 3.33 * 10^5 = 19314 J
Total energy = 2060.16 + 19314 = 21374.16 J

So, we have now 0.058 kg of water at 0˚C, 0.170 kg of water at 26˚ and 21374.16 joules of heat energy.

Let Tf be final temperature.
Q1 = 0.058 *4186 * Tf = 242.79 * Tf
Q2 = 0.170 * 4186 * (26 – Tf) = 18502.12 – 711.62 * Tf
The difference of these is 21374.16
so, 18502.12 – 711.62 * Tf – 242.79 * Tf = 18502.12 – 954.41 * Tf
=> 18502.12 – 954.41 * Tf = 21374.16

=> 954.41 * Tf = 2872.04

=> Tf = 4637.88 ÷ 1280.916
This is approximately 3.0˚C

(b) Q1= 0.029 * 2220 * 16 = 1030.08 J
Q2 = 0.029 * 3.33 * 10^5 = 9657 J
Total = 1030.08 + 9657 = 10687.08 J

Now we have 0.029 kg of water at 0˚C, 0.170 kg of water at 26˚ and 10687.08 J of heat energy. Let Tf be final temperature.

For the 0˚ water, Q1 = 0.029 *4186 * Tf = 121.4 * Tf
For the 26˚ water, Q2 = 18502.12 – 711.62 * Tf
Q2 – Q1 = 18502.12 – 833.02 * Tf

18502.12 – 833.02 * Tf = 10687.08
=> 833.02 * Tf = 7815.04

=> Tf = 9.38 ˚C.