If 125cal of heat is applied to a 60.0-g piece of copper at 20.0(Deg C), what will the final temperature be? The specific heat of copper is 0.0920cal/(g.degC)
125 cal = mass Cu x specific heat Cu x (Tfinal-Tinitial)
To find the final temperature of the copper, we can use the formula:
Q = mcΔT
Where:
Q is the heat transferred to the copper (in calories)
m is the mass of the copper (in grams)
c is the specific heat of copper (in cal/(g.degC))
ΔT is the change in temperature (final temperature - initial temperature) of the copper (in degrees Celsius)
We are given the following values:
Q = 125 cal
m = 60.0 g
c = 0.0920 cal/(g.degC)
T(initial) = 20.0°C
We need to find T(final), the final temperature.
First, let's rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
Next, let's substitute the given values into the rearranged formula:
ΔT = 125 cal / (60.0 g * 0.0920 cal/(g.degC))
Now, we can simplify and calculate ΔT:
ΔT = 2.15 degC
Finally, we can find the final temperature by adding ΔT to the initial temperature:
T(final) = T(initial) + ΔT
T(final) = 20.0°C + 2.15°C
Therefore, the final temperature will be 22.15°C.