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When you take a bath, how many kilograms of hot water (47.1°C) must you mix with...

When you take a bath, how many kilograms of hot water (47.1°C) must you mix with cold water (12.2°C) so that the temperature of the bath is 36.3°C? The total mass of water (hot plus cold) is 188 kg. Ignore any heat flow between the water and its external surroundings.

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Answer #1

Specific heat of water = C = 4186 J/(kg.oC)

Mass of hot water = m1

Mass of cold water = m2

Total mass of water = m = 188 kg

m = m1 + m2

m2 = m - m1

Initial temperature of the hot water = T1 = 47.1 oC

Initial temperature of the cold water = T2 = 12.2 oC

Final temperature of the bath = T3 = 36.3 oC

The heat lost by the hot water is equal to the heat gained by the cold water.

m_{1}C(T_{1} - T_{3}) = m_{2}C(T_{3} - T_{2})

m_{1}(T_{1} - T_{3}) = m_{2}(T_{3} - T_{2})

m_{1}(T_{1} - T_{3}) = (m - m_{1})(T_{3} - T_{2})

m_{1}(47.1 - 36.3) = (188 - m_{1})(36.3 - 12.2)

10.8m_{1} = 4530.8 - 24.1m_{1}

34.9m_{1} = 4530.8

m1 = 129.8 kg

Mass of the hot water = 129.8 kg

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