Calculate how much heat is required to bring 113 g of water from a temperature of 26.3°C to a temperature of 100°C. The specific heat of water is 4186 J/kg∙°C.

Q = m x c x delta t

m = 113 g

c = 4186 J/Kg C = 4.18 J/g C

delta t = 100 - 26.3 = 73.7 degree Celsius

Q = 113 X 4.18 X 73.7

Q = 34811.458 Joules = 3.48 Kilo Joules

Please do not forget to best the solution.

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