QuestionJune 20, 2025

Which of the following substances (with specific heat capacity provided would show the greatest temperature change upon absorbing 100.0 J of heat? 10.0 g Ag, C_(Ag)=0.235J/g^circ C 10.0 g Fe, C_(Fe)=0.449J/g^circ C 10.0 g water, C_(water)=4.18J/g^circ C 10.0 g Ca, C_(Ca)=0.650J/g^circ C 10.0 g Pb, C_(Pb)=0.160J/g^circ C

Which of the following substances (with specific heat capacity provided would show the greatest temperature change upon absorbing 100.0 J of heat? 10.0 g Ag, C_(Ag)=0.235J/g^circ C 10.0 g Fe, C_(Fe)=0.449J/g^circ C 10.0 g water, C_(water)=4.18J/g^circ C 10.0 g Ca, C_(Ca)=0.650J/g^circ C 10.0 g Pb, C_(Pb)=0.160J/g^circ C
Which of the following substances (with specific heat capacity provided would show the greatest temperature change upon absorbing 100.0 J of heat? 10.0 g Ag, C_(Ag)=0.235J/g^circ C 10.0 g Fe, C_(Fe)=0.449J/g^circ C 10.0 g water, C_(water)=4.18J/g^circ C 10.0 g Ca, C_(Ca)=0.650J/g^circ C 10.0 g Pb, C_(Pb)=0.160J/g^circ C

Solution
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Answer

Lead (Pb) with a temperature change of 62.50^{\circ}C shows the greatest temperature change. Explanation 1. Identify the Formula Use the formula for temperature change: \Delta T = \frac{q}{m \cdot C}, where q is heat absorbed, m is mass, and C is specific heat capacity. 2. Calculate Temperature Change for Each Substance For Ag: \Delta T_{Ag} = \frac{100.0 \, J}{10.0 \, g \times 0.235 \, J/g^{\circ}C} = 42.55^{\circ}C For Fe: \Delta T_{Fe} = \frac{100.0 \, J}{10.0 \, g \times 0.449 \, J/g^{\circ}C} = 22.27^{\circ}C For Water: \Delta T_{water} = \frac{100.0 \, J}{10.0 \, g \times 4.18 \, J/g^{\circ}C} = 2.39^{\circ}C For Ca: \Delta T_{Ca} = \frac{100.0 \, J}{10.0 \, g \times 0.650 \, J/g^{\circ}C} = 15.38^{\circ}C For Pb: \Delta T_{Pb} = \frac{100.0 \, J}{10.0 \, g \times 0.160 \, J/g^{\circ}C} = 62.50^{\circ}C

Explanation

1. Identify the Formula<br /> Use the formula for temperature change: $\Delta T = \frac{q}{m \cdot C}$, where $q$ is heat absorbed, $m$ is mass, and $C$ is specific heat capacity.<br /><br />2. Calculate Temperature Change for Each Substance<br /> For Ag: $\Delta T_{Ag} = \frac{100.0 \, J}{10.0 \, g \times 0.235 \, J/g^{\circ}C} = 42.55^{\circ}C$<br /> For Fe: $\Delta T_{Fe} = \frac{100.0 \, J}{10.0 \, g \times 0.449 \, J/g^{\circ}C} = 22.27^{\circ}C$<br /> For Water: $\Delta T_{water} = \frac{100.0 \, J}{10.0 \, g \times 4.18 \, J/g^{\circ}C} = 2.39^{\circ}C$<br /> For Ca: $\Delta T_{Ca} = \frac{100.0 \, J}{10.0 \, g \times 0.650 \, J/g^{\circ}C} = 15.38^{\circ}C$<br /> For Pb: $\Delta T_{Pb} = \frac{100.0 \, J}{10.0 \, g \times 0.160 \, J/g^{\circ}C} = 62.50^{\circ}C$
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