Question

Calculate the least amount of work that must be done to freeze one gram of wate at 0^@C by means of a refrigerator. Temperature of surroundings is 27^@C. How much heat is passed on the surroundings in this process? Latent heat of fusion L=80cal//g.

Answer 1

Explanation:

L=80

g

cal

m=1g

$$T_1=27°=300K$$

$$T_2=0°=273K$$

Least work done W=L×m×

T

2

T

1

=80×1×

273

300

=87.912cal

Answer 2

least amount of work =  7.91 cal.

Heat is passed on the surrounding = 87.91 cal.

Given:

Water temperature at 0°c = 273 k

Temperature of surroundings is 27° c. = 300 k

Latent heat of fusion L= 80cal/g.

To find:

The least amount of work that must be done to freeze.

How much heat is passed on the surroundings in this process.

Formula used:

COP of refrigerator = $\frac{T_2}{T_1 - T_2}$ = $\frac{Heat \ absorbed }{Work \ done}$

$T_1$ = Higher temperature

$T_2$ = lower temperature

Explanation:

There is 1 gm of water and Latent heat of fusion L= 80cal/g.

So Heat absorbed from substance = 1 × 80 = 80  cal

COP of refrigerator = $\frac{T_2}{T_1 - T_2}$

                                = $\frac{273}{300 - 273}$

                                = 10.11

Also,

COP of refrigerator = $\frac{Heat \ absorbed }{Work \ done}$

           Work done   =  $\frac{80}{10.11}$

           Work done   = 7.91 cal

Heat is passed on the surrounding = Heat absorbed + Work done

                                                          = 80 + 7.91

                                                          = 87.91 cal

Note: Refer below attached picture for better understanding.

least amount of work =  7.91 cal

Heat is passed on the surrounding = 87.91 cal

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