Question

A laboratory technician drops a 0.0850 kg sample of unknown material, at a temperature of 100.0∘C, into a calorimeter. The calorimeter can, initially at 19.0∘C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can is 26.1∘C. Compute the specific heat capacity of the sample.

Answer 1

the specific heat capacity of the sample is 1011.05 J /kg k

Step-by-step explanation:

The computation of the specific heat capacity of the sample is shown below:

Given that

Q=mcΔθ

Now Heat gain by the calorimeter is

= (0.200)× (4180) × (26.1 - 19) + (0.15) × (390) ×(26.1 - 19)

=6 350.95J

Now

6350.95=(0.0850) × c ×(100-26.1)

c= 6350.95 ÷ ((0.085) × (100 - 26.1))

= 1011.05 J /kg k

Hence, the specific heat capacity of the sample is 1011.05 J /kg k