QuestionJune 26, 2025

6. How many kilocalories of heat are required to raise the temperature of 6.500 g of ice from -15.0^circ C to -1.5^circ C ? The specific heat of ice is 2.065J/(g^circ C) (4 points)

6. How many kilocalories of heat are required to raise the temperature of 6.500 g of ice from -15.0^circ C to -1.5^circ C ? The specific heat of ice is 2.065J/(g^circ C) (4 points)
6. How many kilocalories of heat are required to raise the temperature of 6.500 g of ice from -15.0^circ C to -1.5^circ C ? The specific heat of ice is 2.065J/(g^circ C) (4 points)

Solution
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Answer

0.042 \, \text{kcal} Explanation 1. Convert specific heat to kilocalories The specific heat of ice is given as 2.065 \, \text{J}/(g^{\circ}C). Convert it to kilocalories using the conversion factor 1 \, \text{cal} = 4.184 \, \text{J} and 1 \, \text{kcal} = 1000 \, \text{cal}. Thus, 2.065 \, \text{J}/(g^{\circ}C) = \frac{2.065}{4.184 \times 1000} \, \text{kcal}/(g^{\circ}C). 2. Calculate temperature change Temperature change \Delta T = -1.5^{\circ}C - (-15.0^{\circ}C) = 13.5^{\circ}C. 3. Calculate heat required Use the formula **q = m \cdot c \cdot \Delta T**, where m = 6.500 \, g, c = \frac{2.065}{4.184 \times 1000} \, \text{kcal}/(g^{\circ}C), and \Delta T = 13.5^{\circ}C. Calculate q.

Explanation

1. Convert specific heat to kilocalories<br /> The specific heat of ice is given as $2.065 \, \text{J}/(g^{\circ}C)$. Convert it to kilocalories using the conversion factor $1 \, \text{cal} = 4.184 \, \text{J}$ and $1 \, \text{kcal} = 1000 \, \text{cal}$. Thus, $2.065 \, \text{J}/(g^{\circ}C) = \frac{2.065}{4.184 \times 1000} \, \text{kcal}/(g^{\circ}C)$.<br /><br />2. Calculate temperature change<br /> Temperature change $\Delta T = -1.5^{\circ}C - (-15.0^{\circ}C) = 13.5^{\circ}C$.<br /><br />3. Calculate heat required<br /> Use the formula **$q = m \cdot c \cdot \Delta T$**, where $m = 6.500 \, g$, $c = \frac{2.065}{4.184 \times 1000} \, \text{kcal}/(g^{\circ}C)$, and $\Delta T = 13.5^{\circ}C$. Calculate $q$.
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