##### A(n) 70-g ice cube at 0°C is placed in 770 g of water at 28°C. What is the final

 Physics Tutor: None Selected Time limit: 1 Day

A(n) 70-g ice cube at 0°C is placed in 770 g of water at 28°C. What is the final temperature of the mixture?

Jun 19th, 2015

Heat of fusion of ice = 334 J/g
Specific heat of water = 4.186 J/g*˚C

1st determine the amount of heat energy required to melt all the ice.
The water’s temperature will decrease as the ice melts.
Energy absorbed by ice = mass * Heat of fusion = 70 * 334 = 23,380Joules

2nd determine the temperature of the water, after releasing the heat energy.

Energy released by water = Mass * Specific heat * ∆T
Energy released by water = 770 * 4.18 * ∆T

770* 4.186 * ∆T = 70 * 334
∆T = 0.138˚
The water’s temperature decreased 0.138˚ as all the ice melted. Final temperature of water = 28 - 9.6 = 27.862˚
Now you have 770 grams of water at 27.862˚C and 70 grams of water at 0˚

3rd determine the final temperature of the water.
70* 4.186 * (Tf – 0) = 770 * 4.186 * (27.862 – Tf)
Divide both sides by 4.186
70* Tf = 770 * (27.862 – Tf)

Tf = 25.54˚

Now, let’s do the entire problem at once to check the answer!

Jun 19th, 2015

...
Jun 19th, 2015
...
Jun 19th, 2015
Dec 3rd, 2016
check_circle