QuestionMay 26, 2025

20. How much heat in calories, is needed to raise the temperature of 125.0 g of Lead (c_(lead)=0.130J/g^circ C) from 17.5^circ C to 41.1^circ C? Use Q=mc^wedge T 4.18 Joules 23.6 Joules 91.7 calories .130J/gC

20. How much heat in calories, is needed to raise the temperature of 125.0 g of Lead (c_(lead)=0.130J/g^circ C) from 17.5^circ C to 41.1^circ C? Use Q=mc^wedge T 4.18 Joules 23.6 Joules 91.7 calories .130J/gC
20. How much heat in calories, is needed to raise the temperature of 125.0 g of Lead
(c_(lead)=0.130J/g^circ C) from 17.5^circ C to 41.1^circ C? Use Q=mc^wedge T
4.18 Joules
23.6 Joules
91.7 calories
.130J/gC

Solution
4.0(180 votes)

Answer

91.7 calories Explanation 1. Identify the formula Use the formula for heat transfer: **Q = mc\Delta T**. 2. Calculate \Delta T \Delta T = 41.1^{\circ}C - 17.5^{\circ}C = 23.6^{\circ}C. 3. Convert specific heat to calories c_{lead} = 0.130 \, \text{J/g}^{\circ}\text{C}; convert to calories: 1 \, \text{cal} = 4.184 \, \text{J}, so c_{lead} = \frac{0.130}{4.184} \approx 0.0311 \, \text{cal/g}^{\circ}\text{C}. 4. Calculate heat in calories Q = (125.0 \, \text{g})(0.0311 \, \text{cal/g}^{\circ}\text{C})(23.6^{\circ}C) \approx 91.7 \, \text{cal}.

Explanation

1. Identify the formula<br /> Use the formula for heat transfer: **$Q = mc\Delta T$**.<br /><br />2. Calculate $\Delta T$<br /> $\Delta T = 41.1^{\circ}C - 17.5^{\circ}C = 23.6^{\circ}C$.<br /><br />3. Convert specific heat to calories<br /> $c_{lead} = 0.130 \, \text{J/g}^{\circ}\text{C}$; convert to calories: $1 \, \text{cal} = 4.184 \, \text{J}$, so $c_{lead} = \frac{0.130}{4.184} \approx 0.0311 \, \text{cal/g}^{\circ}\text{C}$.<br /><br />4. Calculate heat in calories<br /> $Q = (125.0 \, \text{g})(0.0311 \, \text{cal/g}^{\circ}\text{C})(23.6^{\circ}C) \approx 91.7 \, \text{cal}$.
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