QuestionMay 15, 2025

9. A 50.0 mathrm(~g) sample of tungsten loses 5,687 mathrm(~J) of energy tends up with a temperature of 145.0^circ mathrm(C) . What was its initial temperature?

9. A 50.0 mathrm(~g) sample of tungsten loses 5,687 mathrm(~J) of energy tends up with a temperature of 145.0^circ mathrm(C) . What was its initial temperature?
9. A 50.0 mathrm(~g) sample of tungsten loses 5,687 mathrm(~J) of energy tends up with a temperature of 145.0^circ mathrm(C) . What was its initial temperature?

Solution
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Answer

T_i = -703.06^{\circ} \mathrm{C} Explanation 1. Identify the formula Use the formula for heat transfer: q = mc\Delta T, where q is the heat energy, m is mass, c is specific heat capacity, and \Delta T is the change in temperature. 2. Rearrange the formula Solve for initial temperature T_i: \Delta T = T_f - T_i \Rightarrow T_i = T_f - \frac{q}{mc}. 3. Substitute known values Given: q = 5687 \, \mathrm{J}, m = 50.0 \, \mathrm{g} = 0.050 \, \mathrm{kg}, T_f = 145.0^{\circ} \mathrm{C}, c = 134 \, \mathrm{J/(kg \cdot K)} (specific heat of tungsten). Calculate: T_i = 145.0 - \frac{5687}{0.050 \times 134}. 4. Perform calculation T_i = 145.0 - \frac{5687}{6.7} = 145.0 - 848.06.

Explanation

1. Identify the formula<br /> Use the formula for heat transfer: $q = mc\Delta T$, where $q$ is the heat energy, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.<br />2. Rearrange the formula<br /> Solve for initial temperature $T_i$: $\Delta T = T_f - T_i \Rightarrow T_i = T_f - \frac{q}{mc}$.<br />3. Substitute known values<br /> Given: $q = 5687 \, \mathrm{J}$, $m = 50.0 \, \mathrm{g} = 0.050 \, \mathrm{kg}$, $T_f = 145.0^{\circ} \mathrm{C}$, $c = 134 \, \mathrm{J/(kg \cdot K)}$ (specific heat of tungsten).<br /> Calculate: $T_i = 145.0 - \frac{5687}{0.050 \times 134}$.<br />4. Perform calculation<br /> $T_i = 145.0 - \frac{5687}{6.7} = 145.0 - 848.06$.
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