1.

A cup of coffee cools from 150^@ F to 144^@ F in I min in a room temperature at 72^@ F. How much time with the coffee take to cool from 110^@ F to 104^@ F in the same room?

Answer»

1.55 min
2.14 min
2.89 min
3.35 min

Solution :Given, in first case,
` T_2 = 150^@ F, T_2 = 144^@ F, t_1 = 1` min androom temperature, `T_0 = 72^@F`
According to Newton's law of cooling,
` mc ( (T_1 - T_2) )/(t_1) = K ( (T_1+ T_2)/(2)- T_0)`
` mc ( (150 - 144)/(1) ) = K ( (150 + 144 )/(2)- 72 )`
6 mc = 75K.....(i)
In second case `T_1 = 110^@ F, T_2 = 104^@F`
TIME ` = t_2 =? `
By Newton's law of cooling, we have
` mc ( (110 - 104)/(t_2) ) = K ( (110 + 104)/(2) - 72) `
` (6mc)/(t_2) = 35k`...(ii)
From Eqs. (i) and (ii), we get
` (6mc)/( (6mc)/(t_2) ) = (75K)/(35K)`
`RARR t_2 = 15/7 = 2.14` min


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