4. How much energy must be absorbed by water with a mass of 0.5 kg in order to raise the temperature from 30^circ C to 65^circ C Note: Water has a specific heat of 4,190J/kg^circ C

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4. How much energy must be absorbed by water with a mass of 0.5 kg in order to raise the temperature from 30^circ C to 65^circ C Note: Water has a specific heat of 4,190J/kg^circ C

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**Question:**How much energy must be absorbed by water with a mass of \(0.5 \mathrm{~kg}\) in order to raise the temperature from \(30^{\circ} \mathrm{C}\) to \(65^{\circ} \mathrm{C}\)? Note: Water has a specific heat of \(4,190 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C}\).**Answer:**【Explanation】: This problem involves calculating the heat energy absorbed by water as its temperature changes. We are provided with the mass of water (\(m = 0.5 \mathrm{~kg}\)), the initial temperature (\(T_{\text{initial}} = 30^{\circ} \mathrm{C}\)), the final temperature (\(T_{\text{final}} = 65^{\circ} \mathrm{C}\)), and the specific heat of water (\(c = 4,190 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C}\)).We can use the heat energy equation (\(Q = mc\Delta T\)), where \(Q\) is the heat energy, \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the temperature change.Substitute the given values:\[Q = (0.5 \mathrm{~kg}) \times (4,190 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C}) \times (65^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C})\]Now, calculate the result:\[Q = (0.5 \times 4,190 \times 35) \mathrm{~J}\]\[Q = 73,325 \mathrm{~J}\]Therefore, the energy that must be absorbed by water is \(73,325 \mathrm{~J}\).**Answer**: \(73,325 \mathrm{~J}\)

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4. How much energy must be absorbed by water with a mass of 0.5 kg in order to raise the temperature from 30^circ C to 65^circ C Note: Water has a specific heat of 4,190J/kg^circ C

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