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```        Hot water of mass 0.35kg at 80degrees celsius is poured into a copper cup which is initially at 25 degrees celsius. The mass of the cup is 0.80kg. Assume that no energy is lost to the surroundings.
Give - Specific Heat Capacity of water = 4200Jkg-1C-1
Specific Heat Capacity of Copper = 400Jkg-1C-1
Question)  Find the initial Temperature of water.

Thanks,
Vishnu```
7 years ago

Rohith Gandhi
24 Points
```										Dear Vishnu,
since no energy is lost to surroundings
heat lost by water = heat gained by copper cup
let final common temperature of the mixture = T
therefore,
0.35X4200X(80-T) = 0.8X400X(T-25)
117600-1470T = 320T-8000
=> 117600+8000 = 1470T+320T
125600=1790T
=>T = 125600/1790 = 70.16 degree C
final common temperature of mixture = 70.16 degree C
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best Vishnu !!!Regards,Askiitians ExpertsRohith Gandhi P
```
7 years ago
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