Question

0.5kg of lemon squash at 30°C is placed in a
frigerator which can remove heat at an average
te of 30 J s-. How long will it take to cool the
lemon squash to 5°C ? Specific heat capacity of
squash = 4200 J kg- K-
​

Answer 1

Answer:

time taken would be

W=P×t

=> t=W/P

=> t=52500/30

= 1750 s

Answer 2

Time taken to cool the lemon squash is 1750 s or 29.167 min .

Given :

Mass of  lemon squash , m = 0.5 kg .

Decrease in temperature , $\Delta T=30-5=25\ K$

( In difference units can be changed from °C to K )

Average heat removed per second , $H=30\ J/s$ .

Specific heat capacity of  squash , S = 4200 J kg- K

​Now , energy loss is , $E=m\times S\times \Delta T$

Putting values in above equation .

We get ,

$E=0.5\times 4200\times 25\\\\E=52500\ J.$

Now , time taken ,

$t=\dfrac{52500\ J}{30\ J/s}\\\\t=1750\ s\\\\t=29.167\ min$

Hence , this is the required solution .

Learn More :

Thermodynamics

https://brainly.in/question/13255297