A 155g sample of an unknown substance was; heated from 25.0°c to 40°c. In the process, the substance absorbed 2085 J of energy. What is the specific heat of the substance?

0.897 J/g°C

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass (M) of substance = 155g

Initial temperature (T1) = 25.0°C

Final temperature (T2) = 40°C

Change is temperature (ΔT) = T2 – T1 = 40°C – 25.0°C = 15°C

Heat Absorbed (Q) = 2085 J

Specific heat capacity (C) of the substance =?

Step 2:

Determination of the specify heat capacity of the substance.

Applying the equation: Q = MCΔT, the specific heat capacity of the substance can be obtained as follow:

Q = MCΔT

C = Q/MΔT

C = 2085 / (155 x 15)

C = 0.897 J/g°C

Therefore, the specific heat capacity of the substance is 0.897 J/g°C

a) the particles of the chair are strongly held together.