Brass is an alloy made from copper & zinc. A 0.55 kg brass sample at 96.7 C is dropped into 2.19kg of water at 4.4C. If the equilibrium temp is 7.0C, what isthe specific heat of wter is 4186 j/kg C.
Answer in units of J/kg C
m₁ =0.55 kg,t₁ =96.7℃
c₂=4186 J/kg•℃, m₂=2.19 kg, t₂=4.4℃
t= 7℃
Q₁=c₁m₁(t₁-t)
Q₂=c₂m₂(t-t₂)
Q₁=Q₂ =>
c₁m₁(t₁-t) = c₂m₂(t-t₂)
Solve for c₁
m₁ =0.55 kg,t₁ =96.7℃
c₂=4186 J/kg•℃, m₂=2.19 kg, t₂=4.4℃
t= 7℃
Q₁=c₁m₁(t₁-t)
Q₂=c₂m₂(t-t₂)
Q₁=Q₂ =>
c₁m₁(t₁-t) = c₂m₂(t-t₂)
Solve for c₁
To find the specific heat of brass, we can use the principle of energy conservation. The heat lost by the brass will be equal to the heat gained by the water:
Heat lost by brass = Heat gained by water
The heat lost by the brass can be calculated using the formula:
Qbrass = mbrass * cbrass * ΔTbrass
Where:
mbrass = mass of brass (0.55 kg)
cbrass = specific heat of brass (unknown)
ΔTbrass = change in temperature of brass (final temperature - initial temperature)
Similarly, the heat gained by the water can be calculated using the formula:
Qwater = mwater * cwater * ΔTwater
Where:
mwater = mass of water (2.19 kg)
cwater = specific heat of water (4186 J/kg C)
ΔTwater = change in temperature of water (equilibrium temperature - initial temperature)
Since the system is in thermal equilibrium, the heat lost by the brass is equal to the heat gained by the water:
Qbrass = Qwater
Substituting the values we know:
mbrass * cbrass * ΔTbrass = mwater * cwater * ΔTwater
0.55 kg * cbrass * (7.0 - 96.7) C = 2.19 kg * 4186 J/kg C * (7.0 - 4.4) C
Simplifying the equation:
0.55 kg * cbrass * (-89.7) C = 2.19 kg * 4186 J/kg C * 2.6 C
Dividing both sides of the equation by -89.7:
0.55 kg * cbrass = (2.19 kg * 4186 J/kg C * 2.6 C) / -89.7
Calculating the right-hand side of the equation:
0.55 kg * cbrass ≈ -216.5 J/kg C
Dividing both sides of the equation by 0.55 kg:
cbrass ≈ -216.5 J/kg C / 0.55 kg
cbrass ≈ -393 J/kg C
Therefore, the specific heat of brass is approximately -393 J/kg C.