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The specific heat of water is 4184  J / kg ·K. By how many joules does the internal energy of 50  g of water change as it is heated from 21 ^∘ C to 37 ^∘ C ? Assume that the expansion of the water is negligible. The heat added to raise the temperature of the water is ΔQ=c m ΔT=(4184  J / kg ·K)(0.050  kg)(16^∘ C)=3.4 ×10^3  J Notice that ΔT in Celsius is equal to ΔT in kelvin. If we ignore the slight expansion of the water, no work was done on the surroundings and so ΔW=0 . Then, the first law,
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