Solved Problem on Heat

A young man bought a ring that claimed to have 9 grams of gold and 1 gram of copper. To prove the information, the boy, a physics student, heated the ring (which had 10 grams of mass) to 520 °C, which he knew was lower than the melting point of the two metals. He placed the ring in a calorimeter with a heat capacity of 20 cal/°C and which contained 80 grams of water at 18°C. The thermal equilibrium was verified at 20 °C. Assuming that the specific heats in the alloy are 0.09 cal/g°C for copper and 0.03 cal/g°C for gold, determine the masses of copper and gold in the ring.

Problem data:

Ring mass: M = 10 g;
Initial ring temperature: t_r = 520 °C;
Specific heat of copper: c_Cu = 0.09 cal/g°C;
Specific heat of gold: c_Au = 0.03 cal/g°C;
Heat capacity of the calorimeter: C = 20 cal/°C;
Initial temperature of the calorimeter: t_c = 18 °C;
Massa Mass of water: m_w = 80 g;
Initial water temperature: t_w = 18 °C;
Equilibrium temperature: t_eq = 20 °C;
Assuming the specific heat of water: c_w = 1 cal/g°C.

Problem diagram:

Figure 1

	m(g)	c(cal/g°C)	t_i(°C)	t_eq(°C)
Calorimeter	20 cal/°C		18	20
water	80	1	18	20
Ring	10	c	520	20

Table 1

where c is the specific heat of the ring.

Note: We do not know the mass and specific heat of the calorimeter, but we do know its heat capacity C, which is given by the product of the mass multiplied by the specific heat \( C=mc=20\;\text{cal/g°C} \).

Solution

Heat lost and gained in each element:

Calorimeter

Heat received by the calorimeter, the heat transfer equation is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mc\left(t_{eq}-t_{0}\right)} \tag{I} \end{gather} \]

The heat capacity is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {C=mc} \tag{II} \end{gather} \]

substituting expression (II) into expression (I)

\[ \begin{gather} Q_{C}=C\left(t_{eq}-t_{0}\right)\\ Q_{C}=20\times (20-18)\\ Q_{C}=20\times 2\\ Q_{C}=40\;\text{cal} \end{gather} \]

Water

Heat received by the water, applying expression (I)

\[ \begin{gather} Q_{w}=80\times 1\times \left(20-18\right)\\ Q_{w}=80\times 2\\ Q_{w}=160\;\text{cal/g°C} \end{gather} \]

Ring

Heat lost by the ring, applying expression (I)

\[ \begin{gather} Q_{r}=10c\times \left(20-520\right)\\ Q_{r}=-500\times 10c\\ Q_{r}=-5000c \end{gather} \]

The sum of the heats gained and lost are equal to zero.

\[ \begin{gather} \sum Q=0\\ Q_{C}+Q_{w}+Q_{r}=0\\ 40+160-5000c=0\\ 5000c=200\\ c-\frac{200}{5000}\\ c=0.04\;\text{cal/g°C} \end{gather} \tag{III} \]

This is the specific heat of the gold and copper metal alloy. The specific heat of the alloy as a function of the specific heats and mass fractions of each of the metals that make up the alloy is given by

\[ \begin{gather} c=c_{Au}\frac{m_{Au}}{M}+c_{Cu}\frac{m_{Cu}}{M} \tag{IV} \end{gather} \]

where M is the total mass of the metal alloy

\[ M=m_{Au}+m_{Cu} \]

substituting the total mass given in the problem and writing the mass of gold as a function of the mass of copper

\[ \begin{gather} 10=m_{Au}+m_{Cu}\\ m_{Au}=10-m_{Cu} \tag{V} \end{gather} \]

substituting the value found in (III), the problem data, and the expression (V) into expression (IV)

\[ 0.04=0.03\times \frac{(10-m_{Cu})}{10}+0.09\times \frac{m_{Cu}}{10} \]

multiplying both sides of the equation by 10

\[ \begin{gather} \qquad\qquad 0.04=0.03\times \frac{(10-m_{Cu})}{10}+0.09\times \frac{m_{Cu}}{10}\qquad (\times10)\\ 0.04\times 10=0.03\times \frac{(10-m_{Cu})}{10}\times 10+0.09\times \frac{m_{Cu}}{10}\times 10\\ 0.4=0.03\times (10-m_{Cu})+0.09m_{Cu}\\ 0.4=0.3-0.03m_{Cu}+0.09m_{Cu}\\ 0.06m_{Cu}=0.4-0.3\\ 0.06m_{Cu}=0.1\\ m_{Cu}=\frac{0.1}{0.06} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {m_{Cu}\approx 1.7\;\text{g}} \]

substituting this result into expression (V)

\[ m_{Au}=10-1\times 7 \]

\[ \bbox[#FFCCCC,10px] {m_{Au}\approx 8.3\;\text{g}} \]

Note: Expression (IV) is Kopp's Law or Neumann-Kopp's rule for calculating the specific heat of a metal alloy formed by two elements. Generally, for n elements in an alloy sample

\[ \bbox[#99CCFF,10px] {c=\sum _{i=1}^{n}c_{i}f_{i}\qquad ,\qquad f_{i}=\frac{m_{i}}{m}} \]

where c is the specific heat of the metal alloy, c_i is the specific heat of the i-th element, and f_i is the mass fraction of the i-th element (mass of the element in the alloy divided by the total mass of the sample).

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .