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प्रश्न
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 g−1K−1.
उत्तर
Given,
mass (m) = 0.5 kg
Change in temperature = (30 − 5)°C = 25°C = 25 K.
Specific heat capacity of squash = 4200 J g−1K−1
ΔQ = mcΔT
ΔQ = 0.5 × 4200 × 25 = 52500 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" = (Δ"Q")/"P"`
= `1/30 xx 525001`
`= 152500 / 30`
= 1750 s
t = 29 min 10 sec
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