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Suppose you needed to cool off 19.8 grams of water at 33.3 °C to 28 °C. Calculate the energy released in this process.

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Answer: 439.07 J

Step-by-step explanation:

The energy released can be calculated using the formula Q = mcΔT; here Q is the heat energy transferred (in joules), m is the mass of the liquid being cooled (in grams), c is the specific heat capacity of the liquid (joule per gram degrees Celsius), and ΔT is the change in temperature.

since m = 19.8 g

c = 4.184 J/g ·°C

ΔT = 33.3 °C - 28 °C = 5.3 °C

⇒ Q = (19.8 g) (4.184 J/g ·°C)(5.3 °C)

Q = 439.07 J

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