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Multiple Choice
A 2.54 g lead weight, initially at 10.3 °C, is submerged in 7.47 g of water at 52.5 °C in an insulated container. What is the final temperature of both substances at thermal equilibrium? Express the temperature in degrees Celsius.
A
12.7 °C
B
52.5 °C
C
48.2 °C
D
50.1 °C
Verified step by step guidance
1
Identify the concept: This problem involves thermal equilibrium, where two substances exchange heat until they reach the same temperature. The heat lost by the warmer substance equals the heat gained by the cooler substance.
Write the heat transfer equation: Use the formula \( q = mc\Delta T \) for both substances, where \( q \) is the heat exchanged, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
Set up the equation for thermal equilibrium: \( m_{\text{lead}}c_{\text{lead}}(T_f - T_{\text{initial, lead}}) = -m_{\text{water}}c_{\text{water}}(T_f - T_{\text{initial, water}}) \). Here, \( T_f \) is the final temperature, and the negative sign indicates heat loss by water.
Substitute known values: Use the given masses, initial temperatures, and specific heat capacities (\( c_{\text{lead}} = 0.128 \, \text{J/g°C} \) and \( c_{\text{water}} = 4.18 \, \text{J/g°C} \)) into the equation.
Solve for \( T_f \): Rearrange the equation to isolate \( T_f \) and solve for the final temperature, ensuring both sides of the equation are balanced.