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### 15.6 Example 6

How much heat in calories must be added to transform a block of 3 kg of ice at -5 C, in water at + 22 C? (specific heat of ice = 0.50 cal/g*K, specific heat of water = 1 cal/g*K, and heat of fusion of water = 79.7 cal/g).

Solution with FisicaLab

Select the Thermodynamics group and, inside this, the Calorimetry and gases module. Erase the content of the chalkboard. And add one element Process, one element Applied heat (applied to the element Process), one element Block, one element Change of state solid-liquid and one element Liquid. As show the image below (the yellow arrows are only to indicate the direction of the process):

To the element Applied heat, with the conversion factor to kilocalories, we have:

Q

Q @ cal

The element Process, whose name in this issue is irrelevant, represents the change of the water from its solid state at -5 C until liquid state at 22 C. This element must contain the element Block, which represents the water in solid state, the element Change of state solid-liquid, which represents the change of state of the water, and the element Liquid, which represents the final state of the water. These elements gonna be called Solid, Fusion and Liquid, respectively. Then to the element Process we have:

Name

0

Object 1

Solid

Object 2

Fusion

Object 3

Liquid

Object 4

0

Object 5

0

Now to the element Block, which represents the water in solid state and that we will call Solid, because this name have in the element Process, we have:

Name

Solid

m

3

c

0.5 @ cal/g*K

Ti

-5 @ C

Tf

0 @ C

Now the element Change of state, called Fusion, is:

Name

Fusion

m

3

cf

79.7 @ cal/g

Sense

>

And to the element Liquid, which represents the final state of the water and called Liquid, we have:

Name

Liquid

m

3

c

1 @ cal/g*K

Ti

0 @ C

Tf

22 @ C

Now click in the icon Solve to get the answer:

```
Q = 312.600 kcal ;
Status = success.

```

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