Question

What is the final temperature of the water mixture that results when 40 g of ice that is initially at -10 C is added to 200 g of water that is initially at 60 C? (The specific heat of ice is 0.5); the heat of fusion of water is 80 cal

Answer 1

Calculate the final temperature of a mixture of 190.0 grams of ice initially at -11.5 degrees Celsius and 282.5 grams of water initially at 92.5 degrees Celsius.

Heat energy = mass * specific heat * ∆T
AND
Heat energy = mass * heat of fusion

specific heat of ice = 2.06 J/(g * ˚C)
Heat of fusion if ice = 334 J/g
specific heat of water = 4.18 J/(g * ˚C)

Heat energy will flow from the hot water to the cold ice, until the temperature of the ice = 0. Then heat energy will flow from the water to the ice until all the ice is melted. Then heat energy will flow from the hot water to the cold water, until the temperature of all the water is the same! Since we do not know whether the hot water can melt all the ice, we better determine the temperature of the hot water after the ice is melted!

1st the temperature of the ice increases from -10 to 0
Heat energy = 40 * 2.06 * 10 = 824 J

2nd the ice melts
Heat energy = 40 * 334 = 13360 J
Total heat energy = 824 + 13360 = 14184 J

At this point, all the ice has melted. Now you have 40 grams of water at 0˚. The temperature of the 200 grams of hot water has decreased as the ice warmed up and melted. Let’s determine the temperature of the hot water.

14184 = 200 * 4.18 * ∆ T
14184 = 836 * ∆ T
∆ T = 16.97˚C
The temperature of the hot water = 60 – 16.97 = 43.03˚C

This is the temperature of the 200 grams of hot water

3rd the temperature of the water increases
In the 3rd phase, temperature of the cold water increases as the temperature of hot water decreases, until all the water is at the same temperature.

Now the temperature of the 40 grams of 0˚ water increases and the temperature of the 200 grams of 43.03˚C water decreases, until the temperature of all the water is the same, Tf.

40 * 4.18 * (Tf – 0) = 200 * 4.18 * (43.03 – Tf)
Divide both sides by 4.18

40 * Tf = 200 * (43.03 – Tf)
40 * Tf = 8606 – 200 * Tf
Tf = 35.86˚C

This is the final temperature of the water!