QuestionJuly 15, 2025

What is the minimum number of ice cubes at 0.0^circ C needed to cool 240 mL of coffee at 75^circ C to 0.0^circ C ? Assume each ice cube is 18.0 g of water and the density of the coffee is 1.0g/mL Delta H_(fus)=334J/g C_(H_(2)O,I)=4.18J/gcdot ^circ C Number of ice cubes=[?] Hint:The energy from the coffee cooling melts the ice cubes. Round to the nearest whole ice cube.

What is the minimum number of ice cubes at 0.0^circ C needed to cool 240 mL of coffee at 75^circ C to 0.0^circ C ? Assume each ice cube is 18.0 g of water and the density of the coffee is 1.0g/mL Delta H_(fus)=334J/g C_(H_(2)O,I)=4.18J/gcdot ^circ C Number of ice cubes=[?] Hint:The energy from the coffee cooling melts the ice cubes. Round to the nearest whole ice cube.
What is the minimum number of ice cubes at 0.0^circ C needed to cool 240 mL of coffee at 75^circ C to 0.0^circ C ? Assume each ice cube is 18.0 g of water and the density of the coffee is 1.0g/mL Delta H_(fus)=334J/g C_(H_(2)O,I)=4.18J/gcdot ^circ C Number of ice cubes=[?] Hint:The energy from the coffee cooling melts the ice cubes. Round to the nearest whole ice cube.

Solution
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Answer

13 ice cubes Explanation 1. Calculate the energy needed to cool the coffee The mass of the coffee is 240 \, \text{g} (since density is 1.0 \, \text{g/mL}). Use **q = mc\Delta T** where m = 240 \, \text{g}, c = 4.18 \, \text{J/g} \cdot ^{\circ}C, and \Delta T = 75^{\circ}C - 0^{\circ}C = 75^{\circ}C. Thus, q = 240 \times 4.18 \times 75 = 75240 \, \text{J}. 2. Calculate the energy required to melt one ice cube Each ice cube has a mass of 18.0 \, \text{g}. Use **\Delta H_{fus} = 334 \, \text{J/g}**. Energy to melt one ice cube is 18.0 \times 334 = 6012 \, \text{J}. 3. Determine the number of ice cubes needed Divide the total energy by the energy per ice cube: \frac{75240}{6012} \approx 12.52. Round up to the nearest whole number since partial ice cubes are not possible.

Explanation

1. Calculate the energy needed to cool the coffee<br /> The mass of the coffee is $240 \, \text{g}$ (since density is $1.0 \, \text{g/mL}$). Use **$q = mc\Delta T$** where $m = 240 \, \text{g}$, $c = 4.18 \, \text{J/g} \cdot ^{\circ}C$, and $\Delta T = 75^{\circ}C - 0^{\circ}C = 75^{\circ}C$. Thus, $q = 240 \times 4.18 \times 75 = 75240 \, \text{J}$.<br /><br />2. Calculate the energy required to melt one ice cube<br /> Each ice cube has a mass of $18.0 \, \text{g}$. Use **$\Delta H_{fus} = 334 \, \text{J/g}$**. Energy to melt one ice cube is $18.0 \times 334 = 6012 \, \text{J}$.<br /><br />3. Determine the number of ice cubes needed<br /> Divide the total energy by the energy per ice cube: $\frac{75240}{6012} \approx 12.52$. Round up to the nearest whole number since partial ice cubes are not possible.
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