QuestionMay 22, 2025

31. What amount of heat energy would be necessary to raise the temperature of 100.0 g of water at room temperature (25^circ C) to the boiling point (100^circ C) ? The specific heat of water is 1.0cal/g^circ C.(1000cal=1kcal) 750 kcal 75 kcal 100 kcal 7.5 kcal

31. What amount of heat energy would be necessary to raise the temperature of 100.0 g of water at room temperature (25^circ C) to the boiling point (100^circ C) ? The specific heat of water is 1.0cal/g^circ C.(1000cal=1kcal) 750 kcal 75 kcal 100 kcal 7.5 kcal
31. What amount of heat energy would be necessary to raise the temperature of
100.0 g of water at room temperature (25^circ C) to the boiling point (100^circ C) ? The
specific heat of water is 1.0cal/g^circ C.(1000cal=1kcal)
750 kcal
75 kcal
100 kcal
7.5 kcal

Solution
4.2(281 votes)

Answer

7.5 kcal Explanation 1. Identify the formula Use the formula for heat energy: **Q = mc\Delta T**. 2. Define variables m = 100.0 \, \text{g}, c = 1.0 \, \text{cal/g}^{\circ}C, \Delta T = 100^{\circ}C - 25^{\circ}C = 75^{\circ}C. 3. Calculate heat energy Substitute values into the formula: Q = 100.0 \times 1.0 \times 75 = 7500 \, \text{cal}. 4. Convert to kcal Convert calories to kilocalories: 7500 \, \text{cal} = 7.5 \, \text{kcal}.

Explanation

1. Identify the formula<br /> Use the formula for heat energy: **$Q = mc\Delta T$**.<br />2. Define variables<br /> $m = 100.0 \, \text{g}$, $c = 1.0 \, \text{cal/g}^{\circ}C$, $\Delta T = 100^{\circ}C - 25^{\circ}C = 75^{\circ}C$.<br />3. Calculate heat energy<br /> Substitute values into the formula: $Q = 100.0 \times 1.0 \times 75 = 7500 \, \text{cal}$.<br />4. Convert to kcal<br /> Convert calories to kilocalories: $7500 \, \text{cal} = 7.5 \, \text{kcal}$.
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