Question

substance | specific heat (j/g°c)
aluminum | 0.900
copper | 0.385
iron | 0.450
granite | 0.790
25.0 g of each material has 150 j of energy added. which material has the smallest increase in temperature?
options: aluminum, iron, granite, copper

Explanation:

Step1: Recall the heat formula

The formula relating heat (\(Q\)), mass (\(m\)), specific heat (\(c\)), and temperature change (\(\Delta T\)) is \(Q = mc\Delta T\). We can rearrange it to solve for \(\Delta T\): \(\Delta T=\frac{Q}{mc}\).

Step2: Identify given values

We know that \(Q = 150\space J\), \(m = 25.0\space g\) for each material. So we can calculate \(\Delta T\) for each substance by plugging into the formula.

Step3: Calculate \(\Delta T\) for aluminum

For aluminum, \(c = 0.900\space J/g^\circ C\). Substitute into \(\Delta T=\frac{Q}{mc}\):
\(\Delta T_{aluminum}=\frac{150}{25.0\times0.900}=\frac{150}{22.5}\approx6.67^\circ C\)

Step4: Calculate \(\Delta T\) for copper

For copper, \(c = 0.385\space J/g^\circ C\):
\(\Delta T_{copper}=\frac{150}{25.0\times0.385}=\frac{150}{9.625}\approx15.5^\circ C\)

Step5: Calculate \(\Delta T\) for iron

For iron, \(c = 0.450\space J/g^\circ C\):
\(\Delta T_{iron}=\frac{150}{25.0\times0.450}=\frac{150}{11.25}\approx13.3^\circ C\)

Step6: Calculate \(\Delta T\) for granite

For granite, \(c = 0.790\space J/g^\circ C\):
\(\Delta T_{granite}=\frac{150}{25.0\times0.790}=\frac{150}{19.75}\approx7.60^\circ C\)

Step7: Compare the temperature changes

Now we compare the \(\Delta T\) values:

• Aluminum: \(\approx6.67^\circ C\)
• Copper: \(\approx15.5^\circ C\)
• Iron: \(\approx13.3^\circ C\)
• Granite: \(\approx7.60^\circ C\)

The smallest \(\Delta T\) is for aluminum.

Answer:

aluminum