Question

1)Brewed coffee is often too hot to drink right away. You can cool it with an ice cube, but this dilutes it. Or you can buy a device that will cool your coffee without dilution - a 200 aluminum cylinder that you take from your freezer and place in a mug of hot coffee.

q)If the cylinder is cooled to -20C, a typical freezer temperature, and then dropped into a large cup of coffee (essentially water, with a mass of 500g ) at 85C, what is the final temperature of the coffee?answer in 2 sig figs and answer in Celsius.

2)A 50 ice cube at -10 $^\circ {\rm C}$ is placed in an aluminum cup whose initial temperature is 70 $^\circ {\rm C}$ . The system comes to an equilibrium temperature of 20 $^\circ {\rm C}$ .

q)What is the mass of the cup?2 sig figs and answer in kg.

Answer 1

Concepts and reason

This question is based upon calorimetry principle.

For the final temperature of the coffee apply calorimeter principle and solve for the final temperature.

Again, for the mass of the aluminum cup apply the calorimeter principle. For this first find the heat lost by the cup and equate the heat used up in various process and finally solve of the mass of aluminum cup.

Fundamentals

The heat change $\Delta Q$ of a substance for a given change in temperature is given as follows:

$\Delta Q = mc\left( {{T_2} - {T_1}} \right)$

Here, $m$is the mass of the substance, $c$is its specific heat, ${T_1}$ and ${T_2}$are the initial and final temperature.

The heat change of the ice, if it melts is given as follows:

$\Delta Q = mL$

Here, $L$ is the latent heat of fusion of the ice.

According to calorimetry principle, the heat lost is equal to the heat gained.

$\Delta {Q_{lost}} = \Delta {Q_{gain}}$

(a)

Let T be the final temperature of the system and measured in degree Celsius.

The heat gained by the cylinder is,

$\Delta Q = {m_1}{c_1}\left( {T - {T_1}} \right)$

The heat lost by the coffee is given by following expression:

$\Delta Q = {m_2}{c_2}\left( {{T_2} - T} \right)$

Equate the two expressions as follows:

$\begin{array}{c}\\{m_1}{c_1}\left( {T - {T_1}} \right) = {m_2}{c_2}\left( {{T_2} - T} \right)\\\\T = \frac{{{m_1}{c_1}{T_1} + {m_2}{c_2}{T_2}}}{{{m_1}{c_1} + {m_2}{c_2}}}\\\end{array}$

Substitute $910{\rm{ J/kg}} \cdot {\rm{K}}$ for ${c_1}$ , $4200{\rm{ J/kg}} \cdot {\rm{K}}$ for ${c_2}$ , ${\rm{0}}{\rm{.200 kg}}$ for ${m_1}$ , $0.5{\rm{ kg}}$ for ${m_2}$, $- 20{\rm{ }}^\circ {\rm{C}}$ for ${T_1}$ and ${\rm{85 }}^\circ {\rm{C}}$ for ${T_2}$ as follows:

$\begin{array}{c}\\T = \frac{{{m_1}{c_1}{T_1} + {m_2}{c_2}{T_2}}}{{{m_1}{c_1} + {m_2}{c_2}}}\\\\ = \frac{{\left( {{\rm{0}}{\rm{.200 kg}}} \right)\left( {910{\rm{ J/kg}} \cdot {\rm{K}}} \right)\left( { - 20} \right) + \left( {0.5{\rm{ kg}}} \right)\left( {4200{\rm{ J/kg}} \cdot {\rm{K}}} \right)\left( {{\rm{85}}} \right)}}{{\left( {{\rm{0}}{\rm{.200 kg}}} \right)\left( {910{\rm{ J/kg}} \cdot {\rm{K}}} \right) + \left( {0.5{\rm{ kg}}} \right)\left( {4200{\rm{ J/kg}} \cdot {\rm{K}}} \right)}}\\\\ = 76.6{\rm{ }}^\circ {\rm{C}}\\\end{array}$

(b)

The amount of hat energy lost by the aluminum cup is,

$\Delta {Q_{lost}} = mc\left( {{T_2} - {T_1}} \right)$

Substitute $900{\rm{ J/kg}} \cdot {\rm{K}}$ for $c$ , $20{\rm{ }}^\circ {\rm{C}}$ for ${T_1}$ and ${\rm{70 }}^\circ {\rm{C}}$ for ${T_2}$ as follows:

$\begin{array}{c}\\\Delta {Q_{lost}} = mc\left( {{T_2} - {T_1}} \right)\\\\ = m\left( {900{\rm{ J/kg}} \cdot {\rm{K}}} \right)\left( {70{\rm{ }}^\circ {\rm{C}} - 20{\rm{ }}^\circ {\rm{C}}} \right)\\\\ = \left( {45000{\rm{ J/kg}}} \right)m\\\end{array}$

The amount of heat energy needed to increase the temperature of the ice is,

$\Delta {Q_1} = {m_{ice}}{c_{ice}}\left( {{T_2} - {T_1}} \right)$

Substitute $2090{\rm{ J/kg}} \cdot {\rm{K}}$ for ${c_{ice}}$ , $- 10{\rm{ }}^\circ {\rm{C}}$ for ${T_1}$ , ${\rm{0}}{\rm{.05 kg}}$ for ${m_{ice}}$and ${\rm{0 }}^\circ {\rm{C}}$ for ${T_2}$ as follows:

$\begin{array}{c}\\\Delta {Q_1} = \left( {{\rm{0}}{\rm{.05 kg}}} \right)\left( {2090{\rm{ J/kg}} \cdot {\rm{K}}} \right)\left\{ {0{\rm{ }}^\circ {\rm{C}} - \left( { - 10{\rm{ }}^\circ {\rm{C}}} \right)} \right\}\\\\ = 1045{\rm{ J}}\\\end{array}$

The amount of heat energy needed to change the phase of ice to water at $0^\circ {\rm{C}}$ is,

$\Delta {Q_2} = {m_{ice}}{L_f}$

Substitute ${\rm{3}}{\rm{.33}} \times {\rm{1}}{{\rm{0}}^5}{\rm{ J/kg}}$ for ${L_f}$ and ${\rm{0}}{\rm{.05 kg}}$ for ${m_{ice}}$ as follows:

$\begin{array}{c}\\\Delta {Q_2} = mL\\\\ = \left( {{\rm{0}}{\rm{.05 kg}}} \right)\left( {{\rm{3}}{\rm{.33}} \times {\rm{1}}{{\rm{0}}^5}{\rm{ J/kg}}} \right)\\\\ = 16650{\rm{ J}}\\\end{array}$

The amount of heat energy needed to increase the temperature of the water is,

$\Delta {Q_3} = {m_{ice}}{c_{water}}\left( {{T_2} - {T_1}} \right)$

Substitute ${\rm{4190 J/kg}} \cdot {\rm{K}}$ for ${c_{water}}$ , $0{\rm{ }}^\circ {\rm{C}}$ for ${T_1}$ , ${\rm{0}}{\rm{.05 kg}}$ for ${m_{ice}}$and ${\rm{20 }}^\circ {\rm{C}}$ for ${T_2}$ as follows:

$\begin{array}{c}\\\Delta {Q_3} = mc\left( {{T_2} - {T_1}} \right)\\\\ = \left( {{\rm{0}}{\rm{.05 kg}}} \right)\left( {{\rm{4190 J/kg}} \cdot {\rm{K}}} \right)\left( {{\rm{20 }}^\circ {\rm{C}} - 0^\circ {\rm{C}}} \right)\\\\ = 4190{\rm{ J}}\\\end{array}$

Therefore, the total amount of heat energy required for this transformation is given as follows:

$\begin{array}{c}\\\Delta {Q_{gain}} = \Delta {Q_1} + \Delta {Q_2} + \Delta {Q_3}\\\\ = 1045 + 16650 + 4190{\rm{ J}}\\\\ = 21885{\rm{ J}}\\\end{array}$

According to calorimetry principle,

$\Delta {Q_{lost}} = \Delta {Q_{gain}}$

Substitute $\left( {45000{\rm{ J/kg}}} \right)m$ for $\Delta {Q_{lost}}$and $21885{\rm{ J}}$ for $\Delta {Q_{gain}}$ in the equation $\Delta {Q_{lost}} = \Delta {Q_{gain}}$.

$\begin{array}{c}\\\left( {45000{\rm{ J/kg}}} \right)m{\rm{ }} = 2{\rm{1885 J}}\\\\m = \frac{{2{\rm{1885 J}}}}{{45000{\rm{ J/kg}}}}{\rm{ }}\\\\ = 0.49{\rm{ kg}}\\\end{array}$

Ans: Part a

The final temperature of the system is$76.6{\rm{ }}^\circ {\rm{C}}$.

Part b

The mass of the aluminum cup is$0.49{\rm{ kg}}$.