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? 

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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...

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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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