In a physics experiment, Wendy immersed a copper of mass 300g in 100 degree C in boiling water and left it there for several minutes. She then transferred the cube to 0.2 kg of water at 10 degrees C. The final temperature of the mixture was 21 degrees C. What value did she get for the specific heat capacity of copper?
The sum of the heats gained is zero (one will be negative heat gained, or heat lost).
Masscopper*ccopper*(21-100)+masswater*cw*(21-10)=0
solve for ccopper
To calculate the specific heat capacity of copper, we need to use the formula:
Q = mcΔT
Where:
Q represents the heat transferred
m represents the mass
c represents the specific heat capacity
ΔT represents the change in temperature
In this experiment, we have the following information:
Mass of copper (m1) = 300g = 0.3kg
Mass of water (m2) = 0.2kg
Initial temperature of copper (T1) = 100°C
Initial temperature of water (T2) = 10°C
Final temperature of the mixture (Tf) = 21°C
The first step is to calculate the heat transferred from the copper to the water. We can calculate this by considering the heat gained by the water is equal to the heat lost by the copper:
Q1 = Q2
Using the formula for heat transfer:
m1c1ΔT1 = m2c2ΔT2
Substituting the given values:
(0.3kg)(c1)(Tf - T1) = (0.2kg)(c2)(Tf - T2)
Now, isolate the specific heat capacity of copper (c1):
c1 = [(0.2kg)(c2)(Tf - T2)] / [(0.3kg)(Tf - T1)]
Substituting the known values:
c1 = [(0.2kg)(1 cal/g°C)(21°C - 10°C)] / [(0.3kg)(21°C - 100°C)]
Simplifying the equation further:
c1 = [2 cal/°C] / [-27 cal/°C]
Finally, calculating the specific heat capacity of copper:
c1 = -0.074 cal/g°C
It's important to note that the obtained value for the specific heat capacity of copper (-0.074 cal/g°C) seems unusually negative, which could indicate an error in the calculations or measurements performed during the experiment. Double-checking the calculations and ensuring accurate measurements will help verify the result.