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:

Step1: Recall Heat Transfer Principle

In a closed system, heat lost by the hot substance equals heat gained by the cold substances. The formula for heat transfer is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat, and \( \Delta T \) is temperature change.

Step2: Identify Substances and Their Data

- Mystery substance: \( m_s = 0.0892 \, \text{kg} \), \( T_{s,i} = 99.20^\circ\text{C} \), \( T_f = 23.38^\circ\text{C} \), \( c_s = ? \)
- Iron container: \( m_{Fe} = 0.0950 \, \text{kg} \), \( c_{Fe} = 448 \, \text{J/(kg·°C)} \), \( T_{Fe,i} = 21.01^\circ\text{C} \)
- Water: \( m_w = 0.216 \, \text{kg} \), \( c_w = 4186 \, \text{J/(kg·°C)} \), \( T_{w,i} = 21.01^\circ\text{C} \)

Step3: Calculate Heat Gained by Iron and Water

Heat gained by iron: \( Q_{Fe} = m_{Fe}c_{Fe}(T_f - T_{Fe,i}) \)
\( Q_{Fe} = 0.0950 \times 448 \times (23.38 - 21.01) \)
\( Q_{Fe} = 0.0950 \times 448 \times 2.37 \approx 0.0950 \times 1061.76 \approx 100.8672 \, \text{J} \)

Heat gained by water: \( Q_w = m_wc_w(T_f - T_{w,i}) \)
\( Q_w = 0.216 \times 4186 \times (23.38 - 21.01) \)
\( Q_w = 0.216 \times 4186 \times 2.37 \approx 0.216 \times 9920.82 \approx 2142.9 \, \text{J} \)

Total heat gained: \( Q_{gain} = Q_{Fe} + Q_w \approx 100.8672 + 2142.9 \approx 2243.7672 \, \text{J} \)

Step4: Calculate Heat Lost by Mystery Substance

Heat lost by substance: \( Q_{loss} = m_sc_s(T_{s,i} - T_f) \)
Since \( Q_{loss} = Q_{gain} \), we have:
\( 0.0892 \times c_s \times (99.20 - 23.38) = 2243.7672 \)

Step5: Solve for \( c_s \)

First, calculate \( \Delta T_s = 99.20 - 23.38 = 75.82^\circ\text{C} \)
Then, \( c_s = \frac{2243.7672}{0.0892 \times 75.82} \)
\( 0.0892 \times 75.82 \approx 6.763 \)
\( c_s \approx \frac{2243.7672}{6.763} \approx 331.8 \, \text{J/(kg·°C)} \) (approximate, may adjust for precise calculation)

Wait, let's recalculate more precisely:

\( T_f - T_{i,Fe} = 23.38 - 21.01 = 2.37 \)

\( Q_{Fe} = 0.0950 \times 448 \times 2.37 \)
448*2.37 = 1061.76; 0.095*1061.76 = 100.8672 J

\( Q_w = 0.216 * 4186 * 2.37 \)
4186*2.37 = 9920.82; 0.216*9920.82 = 2142.9 J (exact: 0.216*9920.82 = 2142.9 J)

Total \( Q_{gain} = 100.8672 + 2142.9 = 2243.7672 J \)

\( \Delta T_s = 99.20 - 23.38 = 75.82 \)

\( c_s = 2243.7672 / (0.0892 * 75.82) \)

0.0892*75.82 = 0.0892*75 + 0.0892*0.82 = 6.69 + 0.073144 = 6.763144

2243.7672 / 6.763144 ≈ 331.8 J/(kg·°C) (more accurately, let's do division: 2243.7672 ÷ 6.763144 ≈ 331.8)

Wait, maybe I made a mistake in specific heat values. Let's check standard specific heats:

Iron: 448 J/(kg·°C) (correct), Water: 4186 J/(kg·°C) (correct).

Wait, let's recalculate \( Q_w \):

0.216 kg * 4186 J/(kg·°C) * 2.37 °C

4186 * 2.37 = 4186 * 2 + 4186 * 0.37 = 8372 + 1548.82 = 9920.82

0.216 * 9920.82 = 0.2*9920.82 + 0.016*9920.82 = 1984.164 + 158.73312 = 2142.89712 J

\( Q_{Fe} = 0.095 * 448 * 2.37 \)

448*2.37 = 1061.76; 0.095*1061.76 = 100.8672 J

Total \( Q_{gain} = 2142.89712 + 100.8672 = 2243.76432 J \)

\( Q_{loss} = m_s c_s \Delta T_s = 0.0892 * c_s * (99.20 - 23.38) = 0.0892 * c_s * 75.82 \)

So, \( c_s = 2243.76432 / (0.0892 * 75.82) \)

Calculate denominator: 0.0892 * 75.82 = 0.0892*75 + 0.0892*0.82 = 6.69 + 0.073144 = 6.763144

Then, 2243.76432 / 6.763144 ≈ 331.8 J/(kg·°C) (rounded to reasonable sig figs: let's check sig figs. Masses: 0.0892 (3), 0.0950 (3), 0.216 (3); temps: 99.20 (4), 21.01 (4), 23.38 (4). So maybe 3 sig figs: ~332 J/(kg·°C) or more precise.

Wait, maybe I miscalculated the denominator. Let's do 0.0892 * 75.82:

75.82 * 0.08 = 6.0656; 75.82 * 0.0092 = 0.697544; total = 6.0656 + 0.697544 = 6.763144. Correct.

2243.76432 / 6.763144 ≈ 331.8, which is approximately 332 J/(kg·°C) or 0.332 kJ/(kg·°C).

Answer:

Approximately \( \boldsymbol{332 \, \text{J/(kg·°C)}} \) (or more precisely \( \approx 331.8 \, \text{J/(kg·°C)} \))