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Answer to Question #67813 in Chemistry for Jessica Rose
Question #67813
How much energy (in calories) is needed to warm 75.0 grams of water from 14.4 degrees C to 19.2 degrees C?
Expert's answer
Q = cmΔT
ΔT = 19.2 – 14.4 = 4.8˚C = 4.8 K
C (water) = 4.218 kJ/(kg·K)
Q = 4.218 · 0.075 · 4.8 = 1.52 kJ
ΔT = 19.2 – 14.4 = 4.8˚C = 4.8 K
C (water) = 4.218 kJ/(kg·K)
Q = 4.218 · 0.075 · 4.8 = 1.52 kJ
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