QuestionMarch 20, 2026

A 10 g gold coin is heated from 25^circ C to 50^circ C(C_(Au)is0.13J/g-^circ C) What is the Delta H' -32.5J -25J +25J +32.5J

A 10 g gold coin is heated from 25^circ C to 50^circ C(C_(Au)is0.13J/g-^circ C) What is the Delta H' -32.5J -25J +25J +32.5J
A 10 g gold coin is heated from 25^circ C to 50^circ C(C_(Au)is0.13J/g-^circ C)
What is the Delta H'
-32.5J
-25J
+25J
+32.5J

Solution
4.4(166 votes)

Answer

+32.5\ \mathrm{J} Explanation 1. Identify the formula for heat change Use q = m \cdot C \cdot \Delta T. 2. Calculate \Delta T \Delta T = 50 - 25 = 25 \ ^{\circ} \mathrm{C} 3. Calculate heat change q = 10 \cdot 0.13 \cdot 25 = 32.5\ \mathrm{J} 4. Determine sign of \Delta H' Heating means energy absorbed ⇒ \Delta H' > 0 ⇒ +32.5\ \mathrm{J}.

Explanation

1. Identify the formula for heat change <br /> Use $q = m \cdot C \cdot \Delta T$. <br /><br />2. Calculate $\Delta T$ <br /> $\Delta T = 50 - 25 = 25 \ ^{\circ} \mathrm{C}$ <br /><br />3. Calculate heat change <br /> $q = 10 \cdot 0.13 \cdot 25 = 32.5\ \mathrm{J}$ <br /><br />4. Determine sign of $\Delta H'$ <br /> Heating means energy absorbed ⇒ $\Delta H' > 0$ ⇒ $+32.5\ \mathrm{J}$.
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