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How much heat is released when 88.2 g of water cools from 33.8°C to 12.9°C?

Answer 1

Avery Adams

The specific heat of water is #4.186*J*g^-1*K^-1#. This should have been quoted with the question.

Given that water is a very dense substance, it has a high specific heat as far as specific heats go.

Heat transferred #=# #DeltaTxxC_Pxx"mass#

#=# #-20.9*Kxx88.2*gxx4.186*J*g^-1*K^-1#

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Answer 2

Caleb Adams

To calculate the heat released, you can use the formula:

Q = mcΔT

Where: Q = heat released (in joules) m = mass of the substance (in grams) c = specific heat capacity of the substance (for water, it's approximately 4.18 J/g°C) ΔT = change in temperature (final temperature - initial temperature)

Substituting the given values:

m = 88.2 g c = 4.18 J/g°C ΔT = (12.9°C - 33.8°C) = -21.1°C

Q = (88.2 g) * (4.18 J/g°C) * (-21.1°C) Q ≈ -7847.47 J

Therefore, approximately 7847.47 joules of heat is released when 88.2 g of water cools from 33.8°C to 12.9°C.

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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.