Finding heat lost when waters of different temps. Are mixed

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When 90 grams of water at 75°C is mixed with 90 grams at 25°C in an insulated container, the final temperature reaches 50°C. The hot water loses 25°C, which needs to be converted to calories. Using the equation Q=mcΔT, where the specific heat capacity of water is 1 calorie/gram°C, the heat lost can be calculated. The calculation yields 2250 calories for the heat lost by the hot water. This method effectively demonstrates how to apply specific heat concepts to determine heat loss in mixed water temperatures.
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Homework Statement



90 grams of water at 75°C are mixed with an equal amount of water at 25°C in a completely insulated container. The final temperature of the water is 50°C. How much heat is lost by the hot water?

Homework Equations



The hot water drops to 50 degrees Celsius so it has lost 25 degrees celsius. My answer needs to be in calories though and I'm not sure where to go from here. How do I get 25 degrees Celsius to calories?

The Attempt at a Solution


My attempt is written above.
 
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Well, the heat lost by the water will be measured in units of energy, not temperature as you may think. Do you remember this guy?

Q=mc_{w}\Delta T
 
You need specific heat capacity of water.SpecificHeat capacity of water is 1 calorie/gram °C.So from this you can work out the heat given out by the hot water as its temperature falls.
 
Okay thank you very much! My book is very light on math/ equations, I had never seen that Q=mcT equation before... I took 1 cal/gram x 90 grams x 25 degrees. I got 2250 cal which proved to be correct! Thanks very much
 
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