Physics

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.

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Given,

mass (m) = 0.5 kg

Change in temperature = (30 – 5)°C = 25° C = 25 K

Specific heat capacity of squash (c) = 4200 J kg^-1 K^-1

From relation,

$Q = m \times c \times △T \\[0.5em]$

Substituting the values in the formula above we get,

$Q = 0.5 \times 4200 \times 25 \\[0.5em] \Rightarrow Q = 52,500 J$

Let time taken to remove 52500 J of heat be t.

Now it is given that,

30 J of heat is removed in 1 sec

So, 52500 J of heat ⇒ t = ?

$t = \dfrac{1}{30} \times 52,500 \\[0.5em] \Rightarrow t = 1750 \text{ s} \\[0.5em] \Rightarrow t = 29 \text{ min } 10 \text{ sec} \\[0.5em]$

Hence, time taken = 29 min 10 sec

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