Question

calculate the energy required to heat 484.0 mg of graphite from 10.7 °c to 20.4 °c. assume the specific heat capacity of graphite under these conditions is 0.710 j·g⁻¹·k⁻¹. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert mass from mg to g

Since \(1\space g = 1000\space mg\), then \(m=484.0\space mg=\frac{484.0}{1000}\space g = 0.4840\space g\)

Step2: Calculate the temperature change

\(\Delta T=T_{final}-T_{initial}\). Given \(T_{initial}=10.7^{\circ}C\) and \(T_{final}=20.4^{\circ}C\). Since \(\Delta T\) in \(^{\circ}C\) is the same as \(\Delta T\) in \(K\) (because \(T(K)=T(^{\circ}C)+ 273.15\), so \(\Delta T=T_{2}(K)-T_{1}(K)=(T_{2}(^{\circ}C)+273.15)-(T_{1}(^{\circ}C)+273.15)=T_{2}(^{\circ}C)-T_{1}(^{\circ}C)\)), \(\Delta T=20.4 - 10.7=9.7\space K\)

Step3: Use the heat - transfer formula \(q = mc\Delta T\)

where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change. Substitute \(m = 0.4840\space g\), \(c=0.710\space J\cdot g^{-1}\cdot K^{-1}\), and \(\Delta T = 9.7\space K\) into the formula:
\(q=(0.4840\space g)\times(0.710\space J\cdot g^{-1}\cdot K^{-1})\times(9.7\space K)\)
\(q=0.4840\times0.710\times9.7\space J\)
\(q = 0.4840\times6.887\space J\)
\(q=3.33\space J\)

Answer:

\(3.33\space J\)