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Which mass of water at 85 degrees Celsius must be added to 1kg of water at 35 degrees Celsius to reach a temperature of 45 degrees Celsius?

When hot and cold water are mixed, the hot water sample loses heat, while the colder sample gains it and ultimately the whole solution reaches a new temperature (somewhere between the temperatures of the two samples).


Here, we have an unknown mass of water sample at 85 degrees. Let us assume it has a mass of 'm' g. The final temperature of this sample is 45 degrees Celsius.


The amount of heat lost by this...

When hot and cold water are mixed, the hot water sample loses heat, while the colder sample gains it and ultimately the whole solution reaches a new temperature (somewhere between the temperatures of the two samples).


Here, we have an unknown mass of water sample at 85 degrees. Let us assume it has a mass of 'm' g. The final temperature of this sample is 45 degrees Celsius.


The amount of heat lost by this water sample is given as:


Heat lost = mass of water x specific heat capacity of water x temperature change


= m x C x (85 - 45) = 40 mC J (where C is the specific heat capacity of water)


We also have 1 kg or 1000 g of water sample at 35 degrees Celsius, which when mixed with the hotter sample, reaches a final temperature of 45 degrees C. 


The amount of heat gained by this sample is given as:


heat gained = 1000 x C x (45 - 35) = 10000C J


Now, assuming no heat loss, the amount of heat gained by the colder sample is equal to the amount of heat lost by the hotter sample.


Thus, heat lost = heat gained


or, 40 mC = 10000C


or, m = 10000/40 g = 250 g


Thus, 250 g of water at 85 degrees Celsius when mixed with 1000 g of water at 35 degrees Celsius, results in a water sample at 45 degrees Celsius.


Hope this helps.

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