Question

Calculate the specific heat (J/g∘C) for a 18.5-g sample of tin that absorbs 183 J when temperature increases from 35.0 ∘C to 78.6 ∘C.

Answer 1

$Q=cm \Delta T \ \ \ \ \ \Rightarrow \ \ \ c=\frac{Q}{m\Delta T}\\\\ Q=183J\\ m=18,5g\\ \Delta T=78,6^{o}C-35^{o}C=43,6^{o}C\\\\ c=\frac{183J}{18,5g*43,6^{o}C}\approx0,227\frac{J}{g^{o}C}$

Answer 2

Answer:

0.227 J/g°C

Explanation:

Specific heat is the heat change by one gram of substance if it is subjected to change of 1 degree celsius.

The relation between specific heat and mass and change in temperature is:

Q = heat change = mass X specific heat X change in temperature.

Given:

Q = 183 J

Mass = 18.5g

increase in temperature = 78.6-35 = 43.6 °C

Putting values

183 = 18.5X specific heat X 43.6

Specific heat = $\frac{183}{18.5X43.6}= 0.227$$\frac{J}{g^{0}C}$