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`        A liquid cools from 60℃ to 40℃ in 10 minute what is the temperature of liquid in next 10 minute where temperature is surrounding in 5℃`
2 years ago

```							APPLYING NEWTON’S LAW OF COOLING,(60-40)/10=K((60+40)/2-5)                ...........(1)(40-T)/10= K((40+T)/2-5)                 ...........(2)EQN.(1)/EQN.(2)20/(40-T) =  45/((40+T)/2-5)ON SOLVING THE ABOVE EQN.WE GETT = 300/11NOTE: PLEASE CHECK ANY CALCULATION ERRORS.REGARDSHARDIK BAWEJA
```
2 years ago
```							Dear Amar According to Newton's law of cooling, we havemc (T1 - T2)/t = k [(T1 + T2)/2 - T0)]        ---(1)mc/k = [(T1 + T2)/2 - T0)]/[(T1 - T2)/t]mc/k = [(60+40)/2 -5)]/[(60 - 40)/10]        = 45 / 2 = 22.5Substituting T1 = 40 and T2 = ? and T0 = 5  (all in °C) in Eq. (1), we getmc (40- T2)/10 = k [(40+ T2)/2   -  5)] mc/k = (30 + T2)*5/ (40 -T2)22.5 = (30 + T2) * 5/(40 - T2)22 - 4.5 T2 = 30 + T25.5 T2 = -8T2 = -8/5.5Please check calculations.
```
2 years ago
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