Find the heat released in cooling 10g of water at 100C down to 25C

]To calculate the amount of heat entering or leaving a system, the equation  

Q

=

m

c

Δ

T

is used.

Explanation:

m = mass (in grams)

c = specific heat capacity (J/g°C)

ΔT = change in temperature (°C)

Here, we will use the specific heat capacity for liquid water which is 4.19 J/g°C.

The mass given is 25.0 grams.

As for the change in temperature, I will assume that it start off at room temperature, 25°C.

25

°

C

−

0

°

C

=

25

°

C

Q

=

m

c

Δ

T

Q

=

25

g

r

a

m

s

⋅

4.19

J

g

°

C

⋅

25

°

C

Q

=

2618.75

J

Take into account significant figures and the answer should be

2.6

⋅

10

3

J

Answer link

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