When 90 kJ is removed from a 2-kg copper bar its temperature drops from 200 degrees celcius to 85 degrees celcius what is the specific heat of the iron?
Q = M*C*(delta T)
Solve for C, with
deltaT = -115 C
Q = -90 kJ
M = 2.0 kg
0.391
0.39 kJ/kg degree C
To find the specific heat of the copper bar, we can use the equation:
Q = mcΔT
Where:
Q is the amount of heat transferred,
m is the mass of the copper bar,
c is the specific heat of copper, and
ΔT is the change in temperature.
In this case, we know that 90 kJ of heat is removed from a 2-kg copper bar, causing the temperature to drop from 200 degrees Celsius to 85 degrees Celsius.
First, we need to convert the given units to SI units.
90 kJ = 90,000 J
ΔT = (final temperature - initial temperature)
ΔT = (85°C - 200°C) = -115°C
Now, rearrange the equation to solve for c:
c = Q / (m * ΔT)
Substituting the given values:
m = 2 kg
Q = 90,000 J
ΔT = -115°C
c = 90,000 J / (2 kg * (-115°C))
c ≈ -783.3 J / (kg * °C)
Therefore, the specific heat of the copper bar is approximately -783.3 J / (kg * °C).