Aperson suspects that an unknown liquid is water. he adds 110.0 j of heat to 55.0 g of liquid. the temperature of the liquid increases by 2.1 celsius. i s the liquid water? the specific heat of water is 4.18 j/ g-celsius. explain answer

Explanation:

First thing first, you mistyped the specific heat of water, which should be

c

water

=

4.18

J

g

∘

C

Now, a substance's specific heat tells you how much heat is required to increase the temperature of

1 g

of that substance by

1

∘

C

.

In the case of water, you would need

4.18 J

to increase the temperature of

1 g

of water by

1

∘

C

.

Notice that your sample of water has a mass of

1 g

as well, which means that the only factor that will determine the amount of heat needed will be the difference in temperature.

The equation that establishes a relationshop between heat and change in temperature looks like this

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed

c

- the specific heat of the substance, in your case of water

Δ

T

- the change in temperature, defined as the difference between the final temperature and the initial temperature

Plug in your values and solve for

q

to get

q

=

1.00

g

⋅

4.18

J

g

⋅

∘

C

⋅

(

83.7

−

26.5

)

∘

C

q

=

239.096 J

Rounded to three sig figs, the answer will be

q

=

239 J

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