What will be the equilibrium temperature when a 245g block of copper at 285 degrees C is placed in a 145g calorimeter aluminum cup containing 825g of water at 12 C ?
The sum of the heats gained is zero.
heat added to heat copper+ heat added to water + heat added to alumium cup=0
245*Ccopper*(Tf-285C)+825*Cwater*(Tf-12)+145*Caluminum*(Tf-12)=0
solve for Tf
To find the equilibrium temperature, we need to understand the concept of heat transfer and use the principle of conservation of energy.
The heat transfer in this system can be represented by the equation:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2 + m3 * c3 * ΔT3
Where:
m1 = mass of the copper block = 245g
c1 = specific heat capacity of copper = 0.39 J/g°C (Joules per gram-degree Celsius)
ΔT1 = change in temperature of the copper block = equilibrium temperature - initial temperature of the copper block
m2 = mass of the aluminum cup (calorimeter) = 145g
c2 = specific heat capacity of aluminum = 0.89 J/g°C
ΔT2 = change in temperature of the aluminum cup (calorimeter) = equilibrium temperature - initial temperature of the aluminum cup (usually taken to be room temperature)
m3 = mass of water = 825g
c3 = specific heat capacity of water = 4.18 J/g°C
ΔT3 = change in temperature of water = equilibrium temperature - initial temperature of water
In this case, the initial temperature of the copper block is 285°C, the initial temperature of the water is 12°C, and the initial temperature of the aluminum cup can be taken as room temperature (usually around 25°C).
We can rearrange the equation for the equilibrium temperature:
(m1 * c1 * ΔT1) = (m2 * c2 * ΔT2) + (m3 * c3 * ΔT3)
Substituting the given values:
(245g * 0.39 J/g°C * ΔT1) = (145g * 0.89 J/g°C * (T - 25°C)) + (825g * 4.18 J/g°C * (T - 12°C))
Simplifying the equation will give you the value of the equilibrium temperature, T.