4 kg of lead shot at 91.1o C are poured into 4 kg of water at 29.5o C. Find the final temperature (in oC) of the mixture. Use cwater = 4187 J/kg oC and clead = 128 J/kg oC.
Please i need the answer and equation used to solve for the answer Thanks!
To find the final temperature of the mixture, you can use the principle of conservation of energy. The total heat lost by the lead shot will equal the total heat gained by the water, assuming there is no heat loss to the surroundings.
Let's calculate the heat lost by the lead shot first:
Q1 = mcΔT
Where:
Q1 = heat lost by the lead shot
m = mass of the lead shot (4 kg)
c = specific heat capacity of lead (128 J/kg oC)
ΔT = change in temperature
ΔT = final temperature - initial temperature
= Tf - 91.1 oC
Q1 = 4 kg * 128 J/kg oC * (Tf - 91.1 oC)
Now, let's calculate the heat gained by the water:
Q2 = mcΔT
Where:
Q2 = heat gained by the water
m = mass of the water (4 kg)
c = specific heat capacity of water (4187 J/kg oC)
ΔT = change in temperature
ΔT = final temperature - initial temperature
= Tf - 29.5 oC
Q2 = 4 kg * 4187 J/kg oC * (Tf - 29.5 oC)
Since Q1 = Q2 (as per the principle of conservation of energy), we can equate the two equations:
4 kg * 128 J/kg oC * (Tf - 91.1 oC) = 4 kg * 4187 J/kg oC * (Tf - 29.5 oC)
Now, let's solve for Tf:
4 kg * 128 J/kg oC * Tf - 4 kg * 128 J/kg oC * 91.1 oC = 4 kg * 4187 J/kg oC * Tf - 4 kg * 4187 J/kg oC * 29.5 oC
Rearranging the equation and solving for Tf:
4 kg * 128 J/kg oC * Tf - 4 kg * 4187 J/kg oC * Tf = 4 kg * 4187 J/kg oC * 29.5 oC - 4 kg * 128 J/kg oC * 91.1 oC
(4 kg * 128 J/kg oC - 4 kg * 4187 J/kg oC) * Tf = 4 kg * 4187 J/kg oC * 29.5 oC - 4 kg * 128 J/kg oC * 91.1 oC
(Tf * (4 kg * 128 J/kg oC - 4 kg * 4187 J/kg oC) = 4 kg * 4187 J/kg oC * 29.5 oC - 4 kg * 128 J/kg oC * 91.1 oC
Tf = (4 kg * 4187 J/kg oC * 29.5 oC - 4 kg * 128 J/kg oC * 91.1 oC) / (4 kg * 128 J/kg oC - 4 kg * 4187 J/kg oC)
Now plug in the values and calculate the final temperature (Tf).