Question

In the specific heat capacity experiment, the unknown object was cooled, then placed in a calorimeter that contain water. The temperature of the calorimeter and water was measured using a temperature probe. The data was fitted using natural exponent function: 20.3+(24.7-20.3) exp(-t/0.51) Determine the ambient temperature.

Answer 1

According to newtons law of cooling -OT = k(T-T) K → constant -AT is rate of cooling To is the ambient temperature If initial temperature is Ti then In(T=%) = -1 T=T, + (T; - To Jexp(-kt) Comparing with the given equation T = (20.3) + (24.7 – 20.3)exp(-0.51) To = 20.3 ambient temperature To = 20.3