Questions: The temperature of a sample of copper increased by 23.5°C when 269 J of heat was applied. What is the mass of the sample? Substance Specific heat J /(g · °C) ------ lead 0.128 silver 0.235 copper 0.385 iron 0.449 aluminum 0.903 m= g

The temperature of a sample of copper increased by 23.5°C when 269 J of heat was applied.

What is the mass of the sample?

Substance Specific heat J /(g · °C)
------
lead 0.128
silver 0.235
copper 0.385
iron 0.449
aluminum 0.903

m=

g

Transcript text: The temperature of a sample of copper increased by $23.5^{\circ} \mathrm{C}$ when 269 J of heat was applied. What is the mass of the sample? \begin{tabular}{|c|c|} \hline Substance & Specific heat $\mathrm{J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)$ \\ \hline lead & 0.128 \\ \hline silver & 0.235 \\ \hline copper & 0.385 \\ \hline iron & 0.449 \\ \hline aluminum & 0.903 \\ \hline \end{tabular} \[ m= \] $\square$ g

Solution

Solution Steps

Step 1: Identify the Given Values

We are given the following information:

Heat applied, $ Q = 269 \, \text{J} $
Temperature change, $ \Delta T = 23.5^{\circ} \text{C} $
Specific heat of copper, $ c = 0.385 \, \text{J/g}^{\circ} \text{C} $

Step 2: Use the Formula for Heat Transfer

The formula for heat transfer is: \[ Q = mc\Delta T \] where $ m $ is the mass of the sample, $ c $ is the specific heat, and $ \Delta T $ is the change in temperature.

Step 3: Solve for Mass

Rearrange the formula to solve for mass $ m $: \[ m = \frac{Q}{c\Delta T} \]

Step 4: Substitute the Given Values

Substitute the given values into the equation: \[ m = \frac{269 \, \text{J}}{0.385 \, \text{J/g}^{\circ} \text{C} \times 23.5^{\circ} \text{C}} \]

Step 5: Calculate the Mass

Perform the calculation: \[ m = \frac{269}{0.385 \times 23.5} = \frac{269}{9.0475} \approx 29.74 \, \text{g} \]