Question

If uadd 700 kilo joule of heat to 700 grams of water originally at 70 degree Celsius how much water is left in a container

Answer 1

Answer:

Each degree change for 1 gram of liquid water involves 1 calorie, released in cooling or absorbed in heating.

If X is the final temperature in degrees C, then the hotter water lost this many calories:

C = 100 grams * (70 - X) degrees * 1 calorie / degree gram) = (7000 - 100 X) calories

The cooler water absorbed the same number of calories, so

C = 200 grams * (X - 10) degrees * 1 calorie / degree gram) = 200 X - 2000

So 7000 - 100 X = 200 X - 2000.

Solve for X to get your answer. If you get X = 30 degrees C, your math is the same as mine.

Verify by checking that the 200 grams of cold water will increase in temperature by half the temperature loss of the 100 grams of hotter water.