A 20.00 g sample of metal is warmed to 165˚C in an oil bath. The sample is then transferred to a coffee cup calorimeter that contains 125.0 g of water at 5.0˚C. The final temperature of the water is 8.8˚C.
mrtal water
m=20g m=125g
t1=165c t1=5
c=? t2=8.8
c=4.18j/g.c
qtot=qmetal + qwater
=mct + mct
= (20g)(c)(x-165)+(125g)(8.8-5)(4.18)
please help
It isn't clear what you want? I assume that is you want the specific heat of the metal. Here is your work.
= (20g)(c)(x-165)+(125(8.8-5(4.18)
Just a change or two.
1. Make that = 0.
2. x is 8.8. If the final T is 8.8, that is not only the final T for the water but also the final T for the metal.
3. Solve for c.
To solve this problem, we can use the equation for heat transfer, which is given by:
q = mcΔT
Where:
q is the heat transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance,
ΔT is the change in temperature.
In this case, we have two substances: the metal and the water. Let's start by calculating the heat transferred to the metal (qmetal).
First, we need to find ΔT for the metal. We are given that the initial temperature of the metal is 165˚C, but the final temperature is not given. Let's call the final temperature of the metal x˚C.
So, ΔT for the metal is x˚C - 165˚C = (x - 165)˚C.
Next, we can calculate the heat transferred to the metal:
qmetal = (mass of the metal) * (specific heat capacity of the metal) * (ΔT for the metal)
= (20.00 g) * (c) * (x - 165)˚C
Moving on to the heat transferred to the water (qwater), we can use the same equation:
qwater = (mass of the water) * (specific heat capacity of water) * (ΔT for the water)
= (125.0 g) * (4.18 J/g.˚C) * (8.8˚C - 5.0˚C)
Now, we can set up the equation for the total heat transfer:
qtotal = qmetal + qwater
Substituting the expressions we found for qmetal and qwater:
qtotal = (20.00 g) * (c) * (x - 165)˚C + (125.0 g) * (4.18 J/g.˚C) * (8.8˚C - 5.0˚C)
And finally, we know that the final temperature of the water is 8.8˚C, so the total heat transfer is 0 (since the water did not change temperature):
qtotal = 0
Now we can solve the equation for c by setting qtotal equal to 0 and rearranging the expression:
0 = (20.00 g) * (c) * (x - 165)˚C + (125.0 g) * (4.18 J/g.˚C) * (8.8˚C - 5.0˚C)
Simplifying this equation, we can solve for x and find the final temperature of the metal sample.