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At 1 atm, how much energy is required to heat 83.0 g of H2O(s) at –18.0 °C to H2O(g) at 129.0 °C?

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1 Answer

To solve this question, use the q=mcΔt formula.  Here, m=83g, c=4.18 J/(g*K).  Δt is found by Tfinal-Tinitial, which is 129-(-18)=147 degrees C.  Since the conversion from C to K is only adding 273 to the C, the difference in degrees C is equivalent to degrees K. So:
q=(83g)*(4.18 J/g*K)*(147K)=51000.18J=51kJ