Question

part 2: thermal equilibrium & calorimetry
specific heat capacity for different materials
substance specific heat (j/kg·°c) substance specific heat (j/kg·°c)
aluminum 900 mercury 140
alcohol (ethyl) 2400 sand 800
copper 390 silver 230
glass 840 water 4180
iron or steel 450 wood 1700
lead 130 human body 3470
marble 860

1. a piece of lead at 82.0°c is mixed with 0.112 kg of water and an 0.0875 kg aluminum calorimeter cup both initially at 25.0°c. the final temperature of the system is 56.0°c. what is the mass of the piece of lead?

Explanation:

Step1: Identify Heat Transfer Directions

Lead loses heat (since \( T_{\text{lead, initial}} = 82.0^\circ\text{C} > T_{\text{final}} = 56.0^\circ\text{C} \)), while water and aluminum gain heat (\( T_{\text{water, initial}} = T_{\text{aluminum, initial}} = 25.0^\circ\text{C} < T_{\text{final}} \)). So, \( Q_{\text{lead, lost}} = Q_{\text{water, gained}} + Q_{\text{aluminum, gained}} \).

Step2: Recall Specific Heat Formula

The formula for heat transfer is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat, and \( \Delta T \) is temperature change.

Step3: Define Variables for Each Substance

- For lead: \( m_{\text{lead}} = ? \), \( c_{\text{lead}} = 130 \, \text{J/kg}\cdot^\circ\text{C} \), \( \Delta T_{\text{lead}} = 82.0 - 56.0 = 26.0^\circ\text{C} \)
- For water: \( m_{\text{water}} = 0.112 \, \text{kg} \), \( c_{\text{water}} = 4180 \, \text{J/kg}\cdot^\circ\text{C} \), \( \Delta T_{\text{water}} = 56.0 - 25.0 = 31.0^\circ\text{C} \)
- For aluminum: \( m_{\text{aluminum}} = 0.0875 \, \text{kg} \), \( c_{\text{aluminum}} = 900 \, \text{J/kg}\cdot^\circ\text{C} \), \( \Delta T_{\text{aluminum}} = 56.0 - 25.0 = 31.0^\circ\text{C} \)

Step4: Write Heat Equations

- \( Q_{\text{lead, lost}} = m_{\text{lead}} \cdot c_{\text{lead}} \cdot \Delta T_{\text{lead}} \)
- \( Q_{\text{water, gained}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \)
- \( Q_{\text{aluminum, gained}} = m_{\text{aluminum}} \cdot c_{\text{aluminum}} \cdot \Delta T_{\text{aluminum}} \)

Step5: Set Up Energy Conservation Equation

\( m_{\text{lead}} \cdot 130 \cdot 26.0 = (0.112 \cdot 4180 \cdot 31.0) + (0.0875 \cdot 900 \cdot 31.0) \)

Step6: Calculate Right-Hand Side (RHS)

First, calculate \( Q_{\text{water, gained}} \):
\( 0.112 \cdot 4180 \cdot 31.0 = 0.112 \cdot 129580 = 14512.96 \, \text{J} \)

Then, calculate \( Q_{\text{aluminum, gained}} \):
\( 0.0875 \cdot 900 \cdot 31.0 = 87.5 \cdot 9 \cdot 31.0 = 787.5 \cdot 31.0 = 24412.5 \, \text{J} \)

Sum RHS: \( 14512.96 + 24412.5 = 38925.46 \, \text{J} \)

Step7: Solve for \( m_{\text{lead}} \)

Left-Hand Side (LHS): \( m_{\text{lead}} \cdot 130 \cdot 26.0 = m_{\text{lead}} \cdot 3380 \)

So, \( m_{\text{lead}} = \frac{38925.46}{3380} \approx 11.52 \, \text{kg} \)? Wait, no, wait—wait, specific heat of lead is 130 J/kg·°C? Wait, the table: Lead has specific heat 130? Wait, original table: Lead is 130? Wait, no, in the table, Lead is 130? Wait, the user's table: Lead: 130. Wait, but let's recalculate RHS:

Wait, 0.112 kg water: \( 0.112 \times 4180 \times 31 \):
4180*31 = 129580; 129580*0.112 = 129580*0.1 + 129580*0.012 = 12958 + 1554.96 = 14512.96 J (correct).

Aluminum: 0.0875 kg * 900 J/kg·°C * 31 °C:
0.0875*900 = 78.75; 78.75*31 = 2441.25 J (oh! Wait, I made a mistake earlier: 0.0875*900 is 78.75, not 787.5. Oh no! That was the error. So 78.75*31 = 2441.25 J.

So RHS is 14512.96 + 2441.25 = 16954.21 J.

Now LHS: \( m_{\text{lead}} \times 130 \times (82 - 56) = m_{\text{lead}} \times 130 \times 26 = m_{\text{lead}} \times 3380 \).

Thus, \( m_{\text{lead}} = 16954.21 / 3380 ≈ 5.016 \, \text{kg} \)? Wait, no, wait: 130*26=3380. 16954.21 / 3380 ≈ 5.016? Wait, no, 3380*5=16900, so ~5.02 kg. Wait, but let's check the specific heat again. Wait, the table: Lead is 130? Wait, in the user's table, Lead: 130. Wait, but maybe I misread the mass of aluminum. The problem says: 0.0875 kg aluminum calorimeter cup. So 0.0875 kg. So 0.0875*900=78.75. 78.75*31=2441.25. Then water: 0.112*4180*31=14512.96. Sum: 14512.96 + 2441.25 = 16954.21. Then lead's heat loss: m*130*(82-56)=m*130*26=3380m. So m=16954.21/3380≈5.016 kg, so approximately 5.02 kg. Wait, but let's do the calculation again carefully.

Wait, specific heat of lead: 130 J/kg·°C (from table). Mass of aluminum: 0.0875 kg. Mass of water: 0.112 kg. Initial temp of lead: 82, others:25, final:56.

So \( Q_{\text{lead}} = m_l \times c_l \times (T_{l,i} - T_f) \)
\( Q_{\text{water}} = m_w \times c_w \times (T_f - T_{w,i}) \)
\( Q_{\text{al}} = m_{al} \times c_{al} \times (T_f - T_{al,i}) \)

Since \( Q_{\text{lost}} = Q_{\text{gained}} \), so:

\( m_l \times 130 \times (82 - 56) = m_w \times 4180 \times (56 - 25) + m_{al} \times 900 \times (56 - 25) \)

Calculate each term:

\( 82 - 56 = 26 \)
\( 56 - 25 = 31 \)

So:

Left: \( m_l \times 130 \times 26 = m_l \times 3380 \)

Right: \( 0.112 \times 4180 \times 31 + 0.0875 \times 900 \times 31 \)

Factor out 31:

\( 31 \times (0.112 \times 4180 + 0.0875 \times 900) \)

Calculate inside the parentheses:

0.112*4180 = 468.16
0.0875*900 = 78.75
Sum: 468.16 + 78.75 = 546.91

Then right: 31*546.91 = 16954.21

Thus, \( m_l = 16954.21 / 3380 ≈ 5.016 \, \text{kg} \), so approximately 5.02 kg (or more precisely, let's do the division: 16954.21 ÷ 3380 ≈ 5.016, so ~5.02 kg). Wait, but maybe I made a mistake in specific heat? Wait, the table: Lead is 130? Let me check the table again. The user's table: Lead: 130. Yes. Aluminum: 900. Water: 4180. Correct. So the calculation should be correct now (after fixing the aluminum mass multiplication error).

Answer:

The mass of the lead is approximately \(\boldsymbol{5.02 \, \text{kg}}\) (or more precisely, around 5.0 kg if rounded, but the precise calculation gives ~5.02 kg).