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Shane Macguire Grade: upto college level
`        A body cools down from 50°C to 45°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surrounding. `
3 years ago

Deepak Patra
474 Points
```										Sol. 50°C 45°C 40°C
Let the surrounding temperature be ‘T’°C
Avg. t = 50 + 45/2 = 47.5
Avg. temp. diff. from surrounding
T = 47.5 – T
Rate of fall of temp = 50 – 45/5 = 1 °C/mm
From Network’s Law
1°C/mm = bA * t
⇒ bA = 1/t = 1/47.5 – T …(1)
In second case,
Avg, temp = 40 + 45/2 = 42.5
Avg. temp. diff. from surrounding
t’ = 42.5 – t
rate of fall of temp = 45 – 40/8 = 5/8 °C/mm
From Network’s Law
⇒ 5/B = 1/(47.5 - T) * (42.5 - T)
By C & D [Componendo & Dividendo method]
We find, T = 34.1°C

```
3 years ago
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