Question

How much heat does a 23.0 gram ice cube absorb as its temperature increases from-17.4 °C to 0.0 °C ? The specific heat of ice is 2.108 J/(g C).

Answer 1

Answer:

Explanation:

The specific heat of ice tells you how much heat must be absorbed by

1 g

of ice in order for its temperature to increase by

1

∘

C

.

You know that since

c

=

2.108 J g

−

1

C

−

1

you need to provide

2.108 J

of heat per gram of ice in order to increase the temperature of the sample by

1

∘

C

. This means that in order to increase the temperature of

23.0 g

of ice by

1

∘

C

, you need to provide

23.0

g

⋅

2.108 J

1

g

1

∘

C

=

48.484 J g

−

1

So, you know that for a

23.0-g

sample of ice,

48.484 J

are needed in order to increase its temperature by

1

∘

C

. In your case, you need the temperature of the sample to increase by

Δ

T

=

0.0

∘

C

−

(

−

17.4

∘

C

)

=

17.4

∘

C

This means that you will need

17.4

∘

C

⋅

48.484 J

1

∘

C

=

844 J

−−−−−

I'll leave the answer rounded to three sig figs.

Keep in mind that this much heat is needed in order to get you from ice at

−

17.4

∘

C

to ice at

0.0

∘

C

. If you want the melt the ice to liquid at

0.0

∘

C

, you're going to need to provide it with additional heat.