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Select the correct answer. What is the final temperature of a 272-gram wooden block that starts at 23.4^circ C and loses 759 joules of energy? The specific heat capacity of wood is 1.716 joules/gram degree Celsius. Use the formula Q=mCDelta T A. 7.1^circ C B. 10.9^circ C C. 16.3^circ C D. 39.7^circ C

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Select the correct answer. What is the final temperature of a 272-gram wooden block that starts at 23.4^circ C and loses 759 joules of energy? The specific heat capacity of wood is 1.716 joules/gram degree Celsius. Use the formula Q=mCDelta T A. 7.1^circ C B. 10.9^circ C C. 16.3^circ C D. 39.7^circ C

Select the correct answer. What is the final temperature of a 272-gram wooden block that starts at 23.4^circ C and loses 759 joules of energy? The specific heat capacity of wood is 1.716 joules/gram degree Celsius. Use the formula Q=mCDelta T A. 7.1^circ C B. 10.9^circ C C. 16.3^circ C D. 39.7^circ C

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ElizabethVeteran · Tutor for 9 years

Answer

### A. \( 7.1^\circ \text{C} \)

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## Step 1: Define the formula<br />### The formula for heat transferred is \(Q = m \cdot C \cdot \Delta T \).<br />## Step 2: Rearrange the formula<br />### Solve for $\Delta T$: <br />\[ \Delta T = \frac{Q}{m \cdot C} \]<br />## Step 3: Substitute the given values<br />### Given: \( Q = -759 \, \text{J} \) (energy lost), \( m = 27.2 \, \text{g} \), \( C = 1.716 \, \text{J/g} \cdot ^\circ\text{C} \), \( T_{\text{initial}} = 23.4^\circ \text{C} \):<br />\[ \Delta T = \frac{-759}{27.2 \cdot 1.716} \]<br />## Step 4: Calculate \(\Delta T\)<br />### <br />\[ \Delta T = \frac{-759}{46.6752} \approx -16.26^\circ \text{C} \]<br />## Step 5: Determine the final temperature<br />### <br />\[ T_{\text{final}} = T_{\text{initial}} + \Delta T \]<br />\[ T_{\text{final}} = 23.4^\circ \text{C} + (-16.26^\circ \text{C}) \approx 7.14^\circ \text{C} \]
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