Thermodynamics question

Question

What would be the final temperature of the system if 28.4 g of lead at 136 degrees C is dropped into 24.7 g of water at 5.77 degrees C in an insulated container. The specific heat of lead is 0.128 J/g degrees C.

Answer 1

The heat loss by the lead is equal to the heat gained by the water. You can use the equation,

q_gained = -q_loss

q_gained = m_waterc_waterΔT_water = 24.7g(4.18J/g°C)(T_f - 5.77°C)

q_loss = m_leadc_leadΔT_lead = 28.4g(0.128J/g°C)(T_f - 136°C)

(notice how in the equations, the final temperatures are the same for the water and the lead)

Therefore,

24.7g(4.18J/g°C)(T_f - 5.77°C) = - 28.4g(0.128J/g°C)(T_f - 136°C)

Now, solve for T_f

103.246T_f - 595.729 = -3.64T_f + 494.387

T_f = 10.2°C

Answer 2

The heat that is given off by the lead is absorbed by the water and since the container is insulated, we can assume no heat loss to the environment. So Q_lead = -Q_water (since one is giving off heat and the other is absorbing)

Note that the specific heat of water (which is not given in the problem) is 4.186 J/(g°C)

Q = mCΔT

Q_lead = m_leadC_lead(T_lead-f - T_lead-i) = 28.4(0.128)(T_lead-f - 136) = 3.6352T_f - 494.3872

Q_water = m_waterC_water(T_water-f - T_water-i) = 24.7(4.186)(T_water-f - 5.77) = 103.3942T_f - 596.584534

Since Q_lead = -Q_water then:

3.6352T_f - 494.3872 = -(103.3942T_f - 596.584534)

(note that T_leadf-f = T_water-f since they reach an equilibrium temperature)

Solve for T_f:

3.6352T_f - 494.3872 = -103.3942T_f + 596.584534

107.0294T_f = 1090.971734

T_f = 10.2 °C