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How much heat energy is needed to warm 300 g of water from 20⁰c to 95⁰c? The specific heat of water is 4.18 J/g⁰C. I. 74.05 kj II. 65.9 kj III. 94.05 kj IV. 84 kj
Accepted Answer
The correct answer is III. 94.05 kJ. Here's how to calculate it:
Formula: Q = m
c
ΔT
Q = heat energy (in Joules)
m = mass (in grams)
c = specific heat capacity (in J/g⁰C)
ΔT = change in temperature (in ⁰C)
Calculations:
Q = 300 g
4.18 J/g⁰C
(95⁰C - 20⁰C) = 94,020 J
Convert Joules to Kilojoules: 94,020 J / 1000 J/kJ = 94.02 kJ
Rounding: The closest answer choice is III. 94.05 kJ.
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