A piece of copper of mass 40g at 200°C is placed in a copper calorimeter of mass 60g containing 50g of water at 10°C. Ignoring heat losses, what will be the final steady temperature after stirring.
(Specific heat capacity of copper= 0.4J/g°C, specific heat capacity of water= 4.2J/g°C)?
Please help me with that question i want the answer
The sum of the heats gained is zero.
heatgainedcopper+heat gained calorimeter+heat gained water=0
40*.4*(Tf-200)+60*.4*(Tf-10)+50*4.2*(Tf-10)=0
solve for Tfinal Tf
Wishing it could be solved
To find the final steady temperature after stirring, we need to calculate the amount of heat gained or lost by each component and set them equal to each other.
Let's start with the heat gained or lost by the copper piece. We can use the formula:
Q = mcΔT
Where:
Q = heat gained or lost
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
For the copper piece:
m = 40g
c = 0.4 J/g°C
ΔT = Tf - Ti (Final temperature - Initial temperature)
ΔT = Tf - 200°C
Now, let's calculate the heat gained or lost by the water in the calorimeter. Using the same formula:
For the water:
m = 50g
c = 4.2 J/g°C
ΔT = Tf - Ti (Final temperature - Initial temperature)
ΔT = Tf - 10°C
Since the system reaches thermal equilibrium, the heat gained by the copper is equal to the heat lost by the water:
Qcopper = Qwater
Using the equation:
mcopperΔTcopper = mwaterΔTwater
Substituting the values:
40g * 0.4 J/g°C * (Tf - 200°C) = 50g * 4.2 J/g°C * (Tf - 10°C)
Now we can solve for Tf (the final steady temperature) by simplifying and solving the equation:
16(Tf - 200) = 21(Tf - 10)
16Tf - 3200 = 21Tf - 210
5Tf = 2990
Tf = 2990 / 5
Tf = 598°C
Therefore, the final steady temperature after stirring will be 598°C.