Question

A 39.40 g39.40 g sample of a substance is initially at 22.3 °C.22.3 °C. After absorbing 1899 J1899 J of heat, the temperature of the substance is 141.1 °C.141.1 °C. What is the specific heat (?c) of the substance?

Answer 1

Mass of the substance(m) = 39.40 g

The initial temperature of the substant (Ti) = 22.3 oC

The final temperature of the system(Tf) = 141.1 oC

Heat absorbed(q) = 1899 J

The formula to determine the heat is as follows:

q = m X C x $\Delta$ T

Here, heat is q, mass is m, specific heat capacity is C, and temperature difference is $\Delta$ T.

Rearrange the formula for C as follows:

C = q / (m x $\Delta$ T)

Determine $\Delta$ T as follows:

$\Delta$T = Tf - Ti

$\Delta$T = 141.1 oC - 22.3 oC

$\Delta$T = 118.8 oC

Now,

C = 1899 J / (39.40 g x 118.8 oC)

C = 0.406 J/g oC

Hence, the specific heat of the substance is 0.406 J/g oC. [3 S.F]