QuestionMay 9, 2025

4. How much heat is required to raise the temperature of 2.0 kg of water from 20^circ C to 80^circ C ? (Specific heat water: c=4186J/kgcdot ^circ C) 5. A 0.5 kg piece of aluminum is heated from 30^circ C to 150^circ C (Specific heat of aluminum: c=900J/kgcdot ^circ How much energy is absorbed? 6. How much heat is needed to melt 1.5 kg of ice at 0^circ C ? (Latent heat of fusion for ice: L=334,000J/kg) 7. A 0.2 kg sample of steam condenses at 100^circ C How much energy is released during condensation? ( 2.26times 10^wedge 6J/kg) 8. A 100 g copper block cools from 100^circ C to 25^circ C (Specific heat of copper: c=385J/kgcdot ^circ C) How much h is lost?

4. How much heat is required to raise the temperature of 2.0 kg of water from 20^circ C to 80^circ C ? (Specific heat water: c=4186J/kgcdot ^circ C) 5. A 0.5 kg piece of aluminum is heated from 30^circ C to 150^circ C (Specific heat of aluminum: c=900J/kgcdot ^circ How much energy is absorbed? 6. How much heat is needed to melt 1.5 kg of ice at 0^circ C ? (Latent heat of fusion for ice: L=334,000J/kg) 7. A 0.2 kg sample of steam condenses at 100^circ C How much energy is released during condensation? ( 2.26times 10^wedge 6J/kg) 8. A 100 g copper block cools from 100^circ C to 25^circ C (Specific heat of copper: c=385J/kgcdot ^circ C) How much h is lost?
4. How much heat is required to raise the temperature of 2.0 kg of water from 20^circ C to 80^circ C ? (Specific heat water: c=4186J/kgcdot ^circ C) 5. A 0.5 kg piece of aluminum is heated from 30^circ C to 150^circ C (Specific heat of aluminum: c=900J/kgcdot ^circ How much energy is absorbed? 6. How much heat is needed to melt 1.5 kg of ice at 0^circ C ? (Latent heat of fusion for ice: L=334,000J/kg) 7. A 0.2 kg sample of steam condenses at 100^circ C How much energy is released during condensation? ( 2.26times 10^wedge 6J/kg) 8. A 100 g copper block cools from 100^circ C to 25^circ C (Specific heat of copper: c=385J/kgcdot ^circ C) How much h is lost?

Solution
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Answer

1. 502,320 \, \text{J} ### 2. 54,000 \, \text{J} ### 3. 501,000 \, \text{J} ### 4. 452,000 \, \text{J} ### 5. 2,887.5 \, \text{J} Explanation 1. Calculate heat required for water temperature change Use Q = mc\Delta T. For water, m = 2.0 \, \text{kg}, c = 4186 \, \text{J/kg} \cdot ^{\circ}C, and \Delta T = 80 - 20 = 60 \, ^{\circ}C. Q = (2.0)(4186)(60) 2. Calculate heat absorbed by aluminum Use Q = mc\Delta T. For aluminum, m = 0.5 \, \text{kg}, c = 900 \, \text{J/kg} \cdot ^{\circ}C, and \Delta T = 150 - 30 = 120 \, ^{\circ}C. Q = (0.5)(900)(120) 3. Calculate heat needed to melt ice Use Q = mL. For ice, m = 1.5 \, \text{kg} and L = 334,000 \, \text{J/kg}. Q = (1.5)(334,000) 4. Calculate energy released during steam condensation Use Q = mL. For steam, m = 0.2 \, \text{kg} and L = 2.26 \times 10^6 \, \text{J/kg}. Q = (0.2)(2.26 \times 10^6) 5. Calculate heat lost by copper block Use Q = mc\Delta T. For copper, m = 0.1 \, \text{kg}, c = 385 \, \text{J/kg} \cdot ^{\circ}C, and \Delta T = 100 - 25 = 75 \, ^{\circ}C. Q = (0.1)(385)(75)

Explanation

1. Calculate heat required for water temperature change<br /> Use $Q = mc\Delta T$. For water, $m = 2.0 \, \text{kg}$, $c = 4186 \, \text{J/kg} \cdot ^{\circ}C$, and $\Delta T = 80 - 20 = 60 \, ^{\circ}C$.<br />$$ Q = (2.0)(4186)(60) $$<br /><br />2. Calculate heat absorbed by aluminum<br /> Use $Q = mc\Delta T$. For aluminum, $m = 0.5 \, \text{kg}$, $c = 900 \, \text{J/kg} \cdot ^{\circ}C$, and $\Delta T = 150 - 30 = 120 \, ^{\circ}C$.<br />$$ Q = (0.5)(900)(120) $$<br /><br />3. Calculate heat needed to melt ice<br /> Use $Q = mL$. For ice, $m = 1.5 \, \text{kg}$ and $L = 334,000 \, \text{J/kg}$.<br />$$ Q = (1.5)(334,000) $$<br /><br />4. Calculate energy released during steam condensation<br /> Use $Q = mL$. For steam, $m = 0.2 \, \text{kg}$ and $L = 2.26 \times 10^6 \, \text{J/kg}$.<br />$$ Q = (0.2)(2.26 \times 10^6) $$<br /><br />5. Calculate heat lost by copper block<br /> Use $Q = mc\Delta T$. For copper, $m = 0.1 \, \text{kg}$, $c = 385 \, \text{J/kg} \cdot ^{\circ}C$, and $\Delta T = 100 - 25 = 75 \, ^{\circ}C$.<br />$$ Q = (0.1)(385)(75) $$
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