QuestionMay 7, 2025

A 2.40 kg block of ice is heated with 5820 J of heat.The specific heat of ice is 2.10Jcdot g^-1cdot ^circ C^-1 By how much will its temperature rise , assuming it does not melt? a 15,500,000^circ C b 15,500^circ C C 115^circ C d 1.15^circ C

A 2.40 kg block of ice is heated with 5820 J of heat.The specific heat of ice is 2.10Jcdot g^-1cdot ^circ C^-1 By how much will its temperature rise , assuming it does not melt? a 15,500,000^circ C b 15,500^circ C C 115^circ C d 1.15^circ C
A 2.40 kg block of ice is heated with 5820 J of heat.The specific heat of ice is 2.10Jcdot g^-1cdot ^circ C^-1 By how much will its temperature rise , assuming it does not melt? a 15,500,000^circ C b 15,500^circ C C 115^circ C d 1.15^circ C

Solution
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Answer

1.15^{\circ}C Explanation 1. Convert mass to grams 2.40 \, \text{kg} = 2400 \, \text{g} 2. Use the heat formula Apply Q = mc\Delta T where Q = 5820 \, \text{J}, m = 2400 \, \text{g}, c = 2.10 \, \text{J/g°C}. Solve for \Delta T: \Delta T = \frac{Q}{mc} = \frac{5820}{2400 \times 2.10} 3. Calculate temperature rise \Delta T = \frac{5820}{5040} = 1.15^{\circ}C

Explanation

1. Convert mass to grams<br /> $2.40 \, \text{kg} = 2400 \, \text{g}$<br />2. Use the heat formula<br /> Apply $Q = mc\Delta T$ where $Q = 5820 \, \text{J}$, $m = 2400 \, \text{g}$, $c = 2.10 \, \text{J/g°C}$. Solve for $\Delta T$: <br /> $\Delta T = \frac{Q}{mc} = \frac{5820}{2400 \times 2.10}$<br />3. Calculate temperature rise<br /> $\Delta T = \frac{5820}{5040} = 1.15^{\circ}C$
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