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
To find the specific heat of brass, we need to make use of the principle of conservation of energy.
Specific heat is defined as the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius.
In this case, we have a brass sample and water, and we know the initial temperatures of both substances, as well as the equilibrium temperature they reach.
The heat lost by the brass sample is equal to the heat gained by the water, according to the principle of conservation of energy.
The formula to calculate heat is Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.
Let's calculate the heat lost by the brass sample and the heat gained by the water separately, and then equate them to find the specific heat of brass.
Heat lost by the brass = Q1 = m1 * c1 * ΔT1
Heat gained by the water = Q2 = m2 * c2 * ΔT2
Where:
m1 = mass of the brass sample = 0.55 kg
c1 = specific heat of brass (to be determined)
ΔT1 = change in temperature of the brass = equilibrium temperature - initial temperature of the brass = 7.0°C - 96.7°C = -89.7°C
m2 = mass of water = 2.19 kg
c2 = specific heat of water = 4186 J/kg°C
ΔT2 = change in temperature of the water = equilibrium temperature - initial temperature of the water = 7.0°C - 4.4°C = 2.6°C
Now, we have the following equations:
Q1 = m1 * c1 * ΔT1
Q2 = m2 * c2 * ΔT2
Since the heat lost by the brass sample (Q1) is equal to the heat gained by the water (Q2), we can equate the equations:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Plugging in the given values:
0.55 kg * c1 * (-89.7°C) = 2.19 kg * 4186 J/kg°C * 2.6°C
Dividing both sides of the equation by (-89.7°C) and rearranging:
c1 = (2.19 kg * 4186 J/kg°C * 2.6°C) / (0.55 kg * (-89.7°C))
c1 = 9800 J/kg°C
Therefore, the specific heat of brass is 9800 J/kg°C.