Question

0.5 kg of lemon squash at 30°C is placed in a
refrigerator which can remove heat at an average
rate of 30 J s-1. How long will it take to cool the
lemon squash to 5°C ? Specific heat capacity of
squash = 4200 J kg-1 K-1.​

Answer 1

Answer:Change in temperature= 30 − 5 = 25 K.

Explanation:

Answer 2

The time is 1750 sec.

Explanation:

Given that,

Mass = 0.5 kg

Initial temperature = 5°C

Final temperature = 30°C

Pressure = 30 J/s

Specific heat = 4200 J /kg K

We need to calculate the heat

Using formula of heat

$\Delta Q=mc\Delta T$

$\Delta Q=0.5\times4200\times(303-278)$

$\Delta Q=52500\ J$

We need to calculate the time

Using formula of time

$t=\dfrac{\Delta Q}{P}$

$t=\dfrac{52500}{30}$

$t=1750\ s$

Hence, The time is 1750 sec.

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Topic : specific heat

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