A $100 \mathrm{~g}$ glass container contains $250 \mathrm{~g}$ of water at $15.0^{\circ} \mathrm{C}$. A $20.0 \mathrm{~g}$ piece of lead at $100{ }^{\circ} \mathrm{C}$ is added to the water in the container. What is the final temperature of the system in ${ }^{\circ} \mathrm{C}$ ? (specific heat of water $=1.00 \mathrm{cal} / \mathrm{g}$ ${ }^{\circ} \mathrm{C}$, specific heat of glass $=0.200 \mathrm{cal} / \mathrm{g}{ }^{\circ} \mathrm{C}$, specific heat of lead $=0.0310 \mathrm{cal} / \mathrm{g}{ }^{\circ} \mathrm{C}$ )

Question

Quizplus · Accepted Answer

To find the final temperature, we use the principle of conservation of energy, assuming no heat is lost to the surroundings. The heat lost by the lead will equal the heat gained by the water and the glass container. Let $T_f$ be the final temperature.Heat lost by lead = Heat gained by water + Heat gained by glass$$m_{\text{Pb}}c_{\text{Pb}}(T_{\text{initial, Pb}} - T_f) = m_{\text{H2O}}c_{\text{H2O}}(T_f - T_{\text{initial, H2O}}) + m_{\text{glass}}c_{\text{glass}}(T_f - T_{\text{initial, glass}})$$Given:- $m_{\text{Pb}} = 20.0 \, \text{g}$, $c_{\text{Pb}} = 0.0310 \, \text{cal/g}^\circ\text{C}$, $T_{\text{initial, Pb}} = 100^\circ\text{C}$- $m_{\text{H2O}} = 250 \, \text{g}$, $c_{\text{H2O}} = 1.00 \, \text{cal/g}^\circ\text{C}$, $T_{\text{initial, H2O}} = 15.0^\circ\text{C}$- $m_{\text{glass}} = 100 \, \text{g}$, $c_{\text{glass}} = 0.200 \, \text{cal/g}^\circ\text{C}$, $T_{\text{initial, glass}} = 15.0^\circ\text{C}$Plugging in the values and solving for $T_f$ gives a final temperature of approximately $15.2^\circ\text{C}$. This calculation involves algebraic manipulation and solving for $T_f$, which yields a result that matches choice B.

A 100 g100 \mathrm{~g}100 g Glass Container Contains 250 g250 \mathrm{~g}250 g Of Water At 15.0∘C15.0^{\circ} \mathrm{C}15.0∘C

A $100 \mathrm{~g}$ Glass Container Contains $250 \mathrm{~g}$ Of Water At $15.0^{\circ} \mathrm{C}$