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What is the number of joules of heat energy released when 20 grams of water is cooled from 293 K to 283 K?

Answer 1

Ellie Andrews

836 J

Use the formula #q = mCΔT#

q = heat absorbed or released, in joules (J) m = mass C = specific heat capacity ΔT = change in temperature

Plug known values into the formula.

The specific heat capacity of water is #4.18 J/g * K#.

#q = 20(4.18)(293 - 283)# #q = 20(4.18)(10)# #q = 836#

836 joules of heat energy are released.

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Answer 2

Cooper Adams

The specific heat capacity of water is 4.18 J/g°C. Using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can calculate Q.

Q = (20 g) * (4.18 J/g°C) * (283 K - 293 K)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.