Question

a chemist carefully measures the amount of heat needed to raise the temperature of a 765.0 g sample of a pure substance from 33.1°c to 50.4°c. the experiment shows that 55.3 kj of heat are needed. what can the chemist report for the specific heat capacity of the substance? be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert heat from kJ to J

Since \(1\space kJ = 1000\space J\), then \(q=55.3\space kJ\times1000 = 55300\space J\)

Step2: Calculate the temperature change \(\Delta T\)

\(\Delta T=T_{final}-T_{initial}\). Given \(T_{initial}=33.1^{\circ}C\) and \(T_{final} = 50.4^{\circ}C\). Also, \(\Delta T\) in \(^{\circ}C\) is the same as \(\Delta T\) in \(K\) (because \(T(K)=T(^{\circ}C)+ 273.15\), so \(\Delta T=T_2 - T_1=(T_2 + 273.15)-(T_1 + 273.15)=T_{2}-T_{1}\)).
\(\Delta T=50.4 - 33.1=17.3\space K\)

Step3: Use the formula \(q = mc\Delta T\) to solve for \(c\)

Rearrange \(c=\frac{q}{m\Delta T}\). Given \(m = 765.0\space g\), \(q = 55300\space J\), \(\Delta T=17.3\space K\)
\(c=\frac{55300\space J}{765.0\space g\times17.3\space K}\)
First, calculate \(765.0\times17.3 = 765.0\times(17 + 0.3)=765.0\times17+765.0\times0.3=13005+229.5 = 13234.5\)
Then \(c=\frac{55300}{13234.5}\approx4.18\space J\cdot g^{-1}\cdot K^{-1}\)

Answer:

\(4.18\space J\cdot g^{-1}\cdot K^{-1}\)