If you combine 290.0 mL of water at 25.00 °C and 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

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When we mix two liquids at different temperatures, one of them will lose heat and the other will gain it. The final solution will have a temperature somewhere in between the temperatures of the two mixing liquids. In this case, one liquid is at 25 degree Celsius, while the other...

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When we mix two liquids at different temperatures, one of them will lose heat and the other will gain it. The final solution will have a temperature somewhere in between the temperatures of the two mixing liquids. In this case, one liquid is at 25 degree Celsius, while the other liquid is at 95 degree Celsius. Hence, the first liquid will gain heat, while the second one will lose it. Assuming this is a closed system, the amount of heat lost by one liquid is equal to the heat gained by the other.

Let the final temperature of the mixture be T degrees Celsius.

Amount of heat gained by 290 ml water = mass x specific heat capacity x change in temperature 

 = 290 ml x 1 g/ml x specific heat x (T-25)

Amount of heat lost by the other fluid = 100 ml x 1 g/ml x specific heat x (95-T)

Since the amount of heat lost and gained is equal, we get

290 x specific heat x (T-25) = 100 x specific heat x (95-T)

Solving the equation, we get, T = 42.95 degree Celsius.

Hope this helps.

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