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How much water at 100 deg must be added to 1 kg of water at 10 deg to give a final temp of 37deg?
The heat flow will balance according to the equation:
(mass1)(delta T1)(Cp) = (mass2)(delta T2)(Cp)
Cp is specific heat and cancels out because it is the same for both liquids (water).
Mass1 = x (to be solved)
Mass2 = 1.0 kg.
Delta T1 = 100 - 37 = 63 degrees temperature differential
Delta T2 = 37 - 10 = 27 degrees temperature differential
Therefore:
x(63) = 1.0(27)
x = 27/63 = 0.42857 kg = (428.57 grammes)
So 0.42857 kg of 100 deg water added to 1.0 kg of 10 deg water will balance at 37 deg.
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