Question

In the specific heat capacity experiment, the unknown object was cooled, then placed in a calorimeter that contains water. The temperature of the calorimeter and water was measured using a temperature probe. The data was fitted using natural exponent function: 20.9+(22.9-20.9) exp(-t/0.66) Determine the initial temperature of the calorimeter and water mixture.

Answer 1

Fall in Temperature of calorimeter and water with time is given as-

T = 20.9+ (22.9 – 20.9) exp(-t/0.66)

Put t=0 to find initial temperature of calorimeter and water-

T= 20.9+(22.9 – 20.9 expl-0/0.66)

​​​​​​ (since exp(0)=1)

T= 20.9 + (22.9 – 20.9) x 1

T= 22.9°C