Question

question 10
1 pts
the temperature of a 386.6 mg sample of a newly created insulating material rose from 293.74 k to 315.05 k when 16.04 j of heat was applied. what is the specific heat of this material? report your answer in units of j/(°c · g) to 4 sig figs, but do not include units in your answer.

Explanation:

Step1: Convert mass from mg to g

Since \(1\ g = 1000\ mg\), then \(m=386.6\ mg=\frac{386.6}{1000}g = 0.3866\ g\)

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

\(\Delta T=T_{2}-T_{1}\). In Kelvin, \(\Delta T = 315.05 - 293.74=21.31\ K\). Since \(\Delta T\) in \(^{\circ}C\) is the same as \(\Delta T\) in \(K\) (because \(T(^{\circ}C)=T(K)- 273.15\), so \(\Delta T=T_{2}(^{\circ}C)-T_{1}(^{\circ}C)=(T_{2}(K)-273.15)-(T_{1}(K)-273.15)=T_{2}(K)-T_{1}(K)\)), \(\Delta T = 21.31^{\circ}C\)

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

Rearrange \(q = mc\Delta T\) to \(c=\frac{q}{m\Delta T}\). Substitute \(q = 16.04\ J\), \(m = 0.3866\ g\), and \(\Delta T=21.31^{\circ}C\) into the formula: \(c=\frac{16.04}{0.3866\times21.31}\)

$$c=\frac{16.04}{8.238}\approx1.947$$

Answer:

\(1.947\)