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℃

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 GET
T = 300/11
NOTE: PLEASE CHECK ANY CALCULATION ERRORS.

REGARDS
HARDIK BAWEJA

Dear Amar
 
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 = [(60+40)/2 -5)]/[(60 - 40)/10]
        = 45 / 2 = 22.5
Substituting T1 = 40 and T2 = ? and T0 = 5  (all in °C) in Eq. (1), we get
mc (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 + T2
5.5 T2 = -8
T2 = -8/5.5
Please check calculations.