Question

♥️ $\color{yellow}{50 Points}$ ♥️

✔ A body cools from 80°C to 50°C in 5 min. Calculate the time it takes to cool from 60°C to 30°, the temperature of the surrounding is 20°C.

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Answer 1

A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60°C to 30 °C. The temprature of the surroundings is 20°C. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60°C to 30 °C. The temprature of the surroundings is 20°C.

I hope it will help you

Answer 2

HEYA❤

According to Newton's law of cooling, we have mc (T1 - T2)/t = k [(T1 + T2)/2 - T0)] ---------------(1) mc/k = [(T1 + T2)/2 - T0)]/[(T1 - T2)/t] mc/k = [(80+50)/2 - 20)]/[(80 - 50)/5] = 45/6 Substituting T1 = 60 and T2 = 30 and T0 = 20 (all in °C) in Eq. (1), we get mc (60 - 30)/t = k [(60+ 30)/2 - 20)] 30 mc/t = 25 k t = 30 mc/(25k) = (30/25) x (45/6) = 9 min The body takes 9 min to cool from 60°C to 30 °C.

9 minutes ✔✔