Asked by ******tasha******
How many grams of water at 20.0C are necessary to change 800.0g of water at 90.0C to 50.0C
Answers
Answered by
drwls
You want to lower the temperature of the 800 g of 90 C water by 40 C, to 50 C. The heat loss from the hot water will be 800*50 = 40,000 calories, since the specific heat is C = 1.00 cal/(g*C)
The cold water that you add will also end up at 50 C after mixing, and it will have to lose 40,000 calories.
The mass M of cold water must be such that
M*C*30 = 40,000 calories
M = 1333 g
The cold water that you add will also end up at 50 C after mixing, and it will have to lose 40,000 calories.
The mass M of cold water must be such that
M*C*30 = 40,000 calories
M = 1333 g