(DrBob222 please help me)
What would be the final temperature of the system if 16.6 g of lead at 138 ◦C is dropped into 23.4 g of water at 8.68 ◦C in an insulated container? The specific heat of lead is 0.128 J/g◦C.
Please explain with the numbers plugged in to the equation
[16.6(0.128)(t-138]+[23.4(4.184)(t-8.68)]=0 Solve for t.
Couldn't you have followed the Pb/water problem Bob P worked for you?
To find the final temperature of the system, we can use the principle of heat conservation.
The heat gained or lost by an object can be calculated using the formula:
q = m * c * ΔT
Where:
q - heat gained or lost
m - mass of the substance
c - specific heat capacity of the substance
ΔT - change in temperature
In this case, we have two substances: lead and water. The lead is initially at 138 ◦C and the water is initially at 8.68 ◦C. We want to find the final temperature of the system when the two substances reach thermal equilibrium.
First, let's calculate the heat gained or lost by the lead.
q_lead = m_lead * c_lead * ΔT_lead
Putting the values in:
q_lead = 16.6 g * 0.128 J/g◦C * (T_final - 138 ◦C)
Now, let's calculate the heat gained or lost by the water.
q_water = m_water * c_water * ΔT_water
Putting the values in:
q_water = 23.4 g * 4.18 J/g◦C * (T_final - 8.68 ◦C)
According to the principle of heat conservation, the heat gained by the lead should be equal to the heat lost by the water:
q_lead = q_water
16.6 g * 0.128 J/g◦C * (T_final - 138 ◦C) = 23.4 g * 4.18 J/g◦C * (T_final - 8.68 ◦C)
Simplifying the equation:
2.1248 (T_final - 138) = 97.572 (T_final - 8.68)
2.1248T_final - 294.1728 = 97.572T_final - 846.43936
Moving terms to one side:
2.1248T_final - 97.572T_final = -846.43936 +294.1728
-95.4472T_final = -552.26656
Now, let's solve for T_final:
T_final = (-552.26656) / (-95.4472)
T_final ≈ 5.78 ◦C
Therefore, the final temperature of the system when 16.6 g of lead at 138 ◦C is dropped into 23.4 g of water at 8.68 ◦C is approximately 5.78 ◦C.