Question

35.0 g of copper pellets are removed from a 300 C oven and immediately dropped into 90.0 mL of water at 21.0 C in an insulated cup.

What will the new water temperature be?

Answer 1

Concept and reason

The concept needed to solve this problem is law of calorimetry and equation for heat energy added or removed on heating or cooling a substance respectively.

Initially, write the equation for heat energy removed from copper pellets on cooling them to equilibrium temperature. After that writer the equation for heat energy added to water on heating it to equilibrium temperature. Finally, use the law of calorimetry and solve for the final temperature of the water that is the equilibrium temperature of the mixture.

Fundamentals

The amount of heat added or removed to change the temperature of a substance can be calculated using the following formula:

$Q = mc\left( {{T_{\rm{H}}} - {T_{\rm{L}}}} \right)$

Here, m is the mass of the substance, c is the specific capacity of substance, ${T_{\rm{H}}}$ is the highest temperature of the substance, and ${T_{\rm{L}}}$ is the lowest temperature of the substance in the heating or cooling process.

From the law of calorimetry, the heat lost ${Q_{\rm{H}}}$ by the hot body is equal to the heat gained ${Q_{\rm{C}}}$ by the cold body in calorimetric mixture.

${Q_{\rm{H}}} = {Q_{\rm{C}}}$

The mass M of a liquid can be calculated using the following formula:

$M = \rho V$

Here, $\rho$ is the density of the liquid and V is the volume of the liquid.

The amount of heat added or removed to change the temperature of a substance can be calculated using the following formula:

$Q = mc\left( {{T_{\rm{H}}} - {T_{\rm{L}}}} \right)$

Substitute 35.0 g for m, $0.385{\rm{ J/g}} \cdot ^\circ {\rm{C}}$ for c, $300^\circ {\rm{C}}$ for ${T_{\rm{H}}}$ and T for ${T_{\rm{L}}}$ , and solve for the amount of heat energy ${Q_{\rm{H}}}$ removed from hot copped pellets.

$\begin{array}{c}\\{Q_{\rm{H}}} = \left( {35.0{\rm{ g}}} \right)\left( {0.385{\rm{ J/g}} \cdot ^\circ {\rm{C}}} \right)\left( {300^\circ {\rm{C}} - T} \right)\\\\ = \left( {{\rm{13}}{\rm{.475 J/}}^\circ {\rm{C}}} \right)\left( {300^\circ {\rm{C}} - T} \right)\\\end{array}$

Here, T is the equilibrium temperature that is final temperature of the mixture.

Mass of the water in the insulator cup can be calculated using the following formula:

$M = \rho V$

Substitute $1{\rm{ g/c}}{{\rm{m}}^3}$ for $\rho$ and 90.0 mL for V.

$\begin{array}{c}\\M = \left( {1{\rm{ g/c}}{{\rm{m}}^3}} \right)\left( {90.0{\rm{ mL}}\frac{{1{\rm{ c}}{{\rm{m}}^3}}}{{1{\rm{ mL}}}}} \right)\\\\ = 90.0{\rm{ g}}\\\end{array}$

Substitute 35.0 g for m, $4.186{\rm{ J/g}} \cdot ^\circ {\rm{C}}$ for c, T for ${T_{\rm{H}}}$ and $21.0^\circ {\rm{C}}$ for ${T_{\rm{L}}}$ in the equation $Q = mc\left( {{T_{\rm{H}}} - {T_{\rm{L}}}} \right)$ , and solve for the amount of heat energy ${Q_{\rm{C}}}$ added to the water.

$\begin{array}{c}\\{Q_{\rm{C}}} = \left( {90.0{\rm{ g}}} \right)\left( {4.186{\rm{ J/g}} \cdot ^\circ {\rm{C}}} \right)\left( {T - 21.0^\circ {\rm{C}}} \right)\\\\ = \left( {{\rm{376}}{\rm{.74 J/}}^\circ {\rm{C}}} \right)\left( {T - 21.0^\circ {\rm{C}}} \right)\\\end{array}$

From the law of calorimetry, the heat lost ${Q_{\rm{H}}}$ by the hot body is equal to the heat gained ${Q_{\rm{C}}}$ by the cold body in calorimetric mixture.

${Q_{\rm{H}}} = {Q_{\rm{C}}}$

Substitute $\left( {{\rm{13}}{\rm{.475 J/}}^\circ {\rm{C}}} \right)\left( {300^\circ {\rm{C}} - T} \right)$ for ${Q_{\rm{H}}}$ and $\left( {{\rm{376}}{\rm{.74 J/}}^\circ {\rm{C}}} \right)\left( {T - 21.0^\circ {\rm{C}}} \right)$ for ${Q_{\rm{C}}}$ , and solve for T.

$\begin{array}{c}\\\left( {{\rm{13}}{\rm{.475 J/}}^\circ {\rm{C}}} \right)\left( {300^\circ {\rm{C}} - T} \right) = \left( {{\rm{376}}{\rm{.74 J/}}^\circ {\rm{C}}} \right)\left( {T - 21.0^\circ {\rm{C}}} \right)\\\\300^\circ {\rm{C}} - T = \left( {\frac{{{\rm{376}}{\rm{.74 J/}}^\circ {\rm{C}}}}{{{\rm{13}}{\rm{.475 J/}}^\circ {\rm{C}}}}} \right)\left( {T - 21.0^\circ {\rm{C}}} \right)\\\\300^\circ {\rm{C}} - T = \left( {\frac{{{\rm{376}}{\rm{.74 J/}}^\circ {\rm{C}}}}{{{\rm{13}}{\rm{.475 J/}}^\circ {\rm{C}}}}} \right)\left( {T - 21.0^\circ {\rm{C}}} \right)\\\\300^\circ {\rm{C}} - T = {\rm{27}}{\rm{.958}}\left( {T - 21.0^\circ {\rm{C}}} \right)\\\end{array}$

$\begin{array}{c}\\300^\circ {\rm{C}} - T = {\rm{27}}{\rm{.958}}T - {\rm{587}}^\circ {\rm{C}}\\\\{\rm{28}}{\rm{.958}}T = 300^\circ {\rm{C}} + {\rm{587}}^\circ {\rm{C}}\\\\T = \frac{{{\rm{887}}^\circ {\rm{C}}}}{{{\rm{28}}{\rm{.958}}}}\\\\ = {\rm{30}}{\rm{.6}}^\circ {\rm{C}}\\\end{array}$

Ans:

The new temperature of the water is ${\rm{30}}{\rm{.6}}^\circ {\rm{C}}$ .