Question

Lead has a specific heat of 0.13 J/g *C if 137 J of heat energy needed to raise the temperature of a sample from 25*c to 45*c, what is the mass of the sample.

Answer 1

The mass of the sample is 52.69 grams

Step-by-step explanation

Given data

`\begin{gathered} q\text{ = 137 J} \\ c\text{ = 0.13 J/g}\degree C \\ \theta1\text{ = 25}\degree C \\ \theta2\text{ = 45}\degree C \end{gathered}`

To find the mass of the sample, we will need to apply the below formula

`\begin{gathered} q\text{ = mc \lparen}\theta2\text{ - }\theta1) \\ \end{gathered}`

The next step is to substitute the above data into the above formula

`\begin{gathered} 137\text{ = m }*\text{ 0.13 }*\text{ \lparen45 - 25\rparen} \\ 137\text{ = m }*\text{ 0.13 }*\text{ 20} \\ 137\text{ = m }*\text{ 2.6} \\ 137\text{ = 2.6m} \\ \text{ Divide both sides by 2.6} \\ (137)/(2.6)\text{ = }(2.6m)/(2.6) \\ m\text{ = 52.69 grams} \end{gathered}`

Hence, the mass of the sample is 52.69 grams