Question

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?
2. 0.0892 kg of a mystery substance is at 99.20°c, and it is placed in a 0.0950 kg iron container holding 0.216 kg of water both at 21.01°c. the final temperature is 23.38°c. what is the specific heat of the substance?

Explanation:

Problem 4:

Step1: Recall Heat Transfer Formula

The heat lost by lead equals heat gained by water and aluminum. Formula: \( Q = mc\Delta T \), where \( Q \) is heat, \( m \) is mass, \( c \) is specific heat, \( \Delta T \) is temperature change. Specific heats: \( c_{\text{lead}} = 130 \, \text{J/(kg·°C)} \), \( c_{\text{water}} = 4186 \, \text{J/(kg·°C)} \), \( c_{\text{aluminum}} = 900 \, \text{J/(kg·°C)} \).

Step2: Set Up Heat Balance

Heat lost by lead: \( Q_{\text{lost}} = m_{\text{lead}} c_{\text{lead}} (T_{\text{lead, initial}} - T_{\text{final}}) \)
Heat gained by water: \( Q_{\text{water}} = m_{\text{water}} c_{\text{water}} (T_{\text{final}} - T_{\text{water, initial}}) \)
Heat gained by aluminum: \( Q_{\text{aluminum}} = m_{\text{aluminum}} c_{\text{aluminum}} (T_{\text{final}} - T_{\text{aluminum, initial}}) \)
Equation: \( m_{\text{lead}} c_{\text{lead}} (82.0 - 56.0) = m_{\text{water}} c_{\text{water}} (56.0 - 25.0) + m_{\text{aluminum}} c_{\text{aluminum}} (56.0 - 25.0) \)

Step3: Plug in Values

\( m_{\text{lead}} \times 130 \times 26 = 0.112 \times 4186 \times 31 + 0.0875 \times 900 \times 31 \)
Calculate right side:
\( 0.112 \times 4186 \times 31 = 0.112 \times 129766 = 14533.792 \)
\( 0.0875 \times 900 \times 31 = 0.0875 \times 27900 = 2441.25 \)
Total gain: \( 14533.792 + 2441.25 = 16975.042 \)

Step4: Solve for \( m_{\text{lead}} \)

\( m_{\text{lead}} \times 130 \times 26 = 16975.042 \)
\( m_{\text{lead}} \times 3380 = 16975.042 \)
\( m_{\text{lead}} = \frac{16975.042}{3380} \approx 5.02 \, \text{kg} \) (Wait, unit check: original masses in kg, but 5.02 kg seems large? Wait, maybe typo in specific heat? Wait, lead's specific heat is 130 J/(kg·°C), yes. Wait, 82-56=26, 56-25=31. Wait, maybe the initial answer was 0.502 kg? Wait, maybe I messed up decimal. Wait, 0.112 kg water, 0.0875 kg aluminum. Let me recalculate:

\( 0.112 \times 4186 = 468.832 \); \( 468.832 \times 31 = 14533.792 \)
\( 0.0875 \times 900 = 78.75 \); \( 78.75 \times 31 = 2441.25 \)
Total: 14533.792 + 2441.25 = 16975.042
\( m_{\text{lead}} \times 130 \times 26 = m_{\text{lead}} \times 3380 = 16975.042 \)
\( m_{\text{lead}} = 16975.042 / 3380 ≈ 5.02 \, \text{kg} \)? But lead's density is high, but maybe. Alternatively, maybe specific heat of lead is 130 J/(g·°C)? No, 130 J/(kg·°C) is correct (130 J per kg per °C). So maybe the answer is 5.02 kg (or 0.00502 t, but units in kg, so 5.02 kg).

Problem 5:

Step1: Heat Balance Equation

Heat lost by mystery substance = heat gained by iron and water.
\( m_{\text{mystery}} c_{\text{mystery}} (T_{\text{mystery, initial}} - T_{\text{final}}) = m_{\text{iron}} c_{\text{iron}} (T_{\text{final}} - T_{\text{iron, initial}}) + m_{\text{water}} c_{\text{water}} (T_{\text{final}} - T_{\text{water, initial}}) \)
Specific heat of iron: \( c_{\text{iron}} = 448 \, \text{J/(kg·°C)} \), \( c_{\text{water}} = 4186 \, \text{J/(kg·°C)} \)

Step2: Plug in Values

\( 0.0892 \times c_{\text{mystery}} \times (99.20 - 23.38) = 0.0950 \times 448 \times (23.38 - 21.01) + 0.216 \times 4186 \times (23.38 - 21.01) \)

Step3: Calculate Temperature Changes

\( 99.20 - 23.38 = 75.82 \)
\( 23.38 - 21.01 = 2.37 \)

Step4: Calculate Right Side

Iron: \( 0.0950 \times 448 \times 2.37 = 0.0950 \times 1061.76 = 100.8672 \)
Water: \( 0.216 \times 4186 \times 2.37 = 0.216 \times 9920.82 = 2142.9 \) (approx)
Total gain: \( 100.8672 + 2142.9 ≈ 2243.7672 \)

Step5: Solve for \( c_{\text{mystery}} \)

\( 0.0892 \times c_{\text{mystery}} \times 75.82 = 2243.7672 \)
\( c_{\text{mystery}} \times (0.0892 \times 75.82) = 2243.7672 \)
\( 0.0892 \times 75.82 ≈ 6.76 \)
\( c_{\text{mystery}} ≈ 2243.7672 / 6.76 ≈ 332 \, \text{J/(kg·°C)} \) (matches the 333 input, likely rounding differences)

Problem 4 Answer:

\(\boxed{5.02 \, \text{kg}}\) (or if decimal error, maybe 0.502 kg, but based on calculation, 5.02 kg)

Problem 5 Answer:

\(\boxed{332 \, \text{J/(kg·°C)}}\) (or 333 with rounding)

Answer:

Step1: Heat Balance Equation

Heat lost by mystery substance = heat gained by iron and water.
\( m_{\text{mystery}} c_{\text{mystery}} (T_{\text{mystery, initial}} - T_{\text{final}}) = m_{\text{iron}} c_{\text{iron}} (T_{\text{final}} - T_{\text{iron, initial}}) + m_{\text{water}} c_{\text{water}} (T_{\text{final}} - T_{\text{water, initial}}) \)
Specific heat of iron: \( c_{\text{iron}} = 448 \, \text{J/(kg·°C)} \), \( c_{\text{water}} = 4186 \, \text{J/(kg·°C)} \)

Step2: Plug in Values

\( 0.0892 \times c_{\text{mystery}} \times (99.20 - 23.38) = 0.0950 \times 448 \times (23.38 - 21.01) + 0.216 \times 4186 \times (23.38 - 21.01) \)

Step3: Calculate Temperature Changes

\( 99.20 - 23.38 = 75.82 \)
\( 23.38 - 21.01 = 2.37 \)

Step4: Calculate Right Side

Iron: \( 0.0950 \times 448 \times 2.37 = 0.0950 \times 1061.76 = 100.8672 \)
Water: \( 0.216 \times 4186 \times 2.37 = 0.216 \times 9920.82 = 2142.9 \) (approx)
Total gain: \( 100.8672 + 2142.9 ≈ 2243.7672 \)

Step5: Solve for \( c_{\text{mystery}} \)

\( 0.0892 \times c_{\text{mystery}} \times 75.82 = 2243.7672 \)
\( c_{\text{mystery}} \times (0.0892 \times 75.82) = 2243.7672 \)
\( 0.0892 \times 75.82 ≈ 6.76 \)
\( c_{\text{mystery}} ≈ 2243.7672 / 6.76 ≈ 332 \, \text{J/(kg·°C)} \) (matches the 333 input, likely rounding differences)

Problem 4 Answer:

\(\boxed{5.02 \, \text{kg}}\) (or if decimal error, maybe 0.502 kg, but based on calculation, 5.02 kg)

Problem 5 Answer:

\(\boxed{332 \, \text{J/(kg·°C)}}\) (or 333 with rounding)