A copper calorimeter can with mass 0.202 kg contains 0.185 kg of water and 1.40×10^−2 kg of ice in thermal equilibrium at atmospheric pressure. If 0.900 kg of lead at a temperature of 417 degrees celsius is dropped into the can, what is the final temperature of the system if no heat is lost to the surroundings?
The sum of the heats gained is zero (some will be negative, as in heat lost).
heat gained calorimeter=masscalor*ccu*(Tf-273)
heat gained by water= (masswater+massmelted ice)*cw*(tf-0)
heat gained by ice= Hf(massice)
Heat gained by lead=masslead*clead*(tf-(417+273))
add all these, set to zero, and solve for Tf.
I'm putting in the answer I get but the website that I do the homework on says the answer is wrong. I'm getting 94.77 K for final temperature, but it says that's wrong. What should I do?
I would recheck it. You know it cant be 94Kelvin. That is pretty cold.
To find the final temperature of the system, we can use the principle of conservation of energy.
First, let's calculate the heat gained or lost by each component in the system.
1. Heat gained by the copper calorimeter (Qc):
The formula for heat gained or lost is Q = m * c * ∆T, where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ∆T is the change in temperature.
The specific heat capacity of copper is approximately 387 J/kg*K, and the mass of the copper calorimeter is 0.202 kg. Since the final temperature is not known, we'll call it Tf.
Qc = mc * cc * (Tf - Ti)
Here, Ti is the initial temperature of the water and ice, which we assume to be 0 degrees Celsius.
2. Heat gained by the water (Qw):
The specific heat capacity of water is 4186 J/kg*K, and the mass of the water is 0.185 kg.
Qw = mw * cw * (Tf - Ti)
3. Heat gained by the ice (Qi):
The heat required to melt the ice into water is given by the formula Qm = m * Lf, where Qm is the heat required, m is the mass of ice, and Lf is the latent heat of fusion of ice (approximately 334,000 J/kg).
Qi = mi * Lf
4. Heat gained by the lead (Ql):
The specific heat capacity of lead is 128 J/kg*K, and the mass of lead is 0.900 kg. The initial temperature of the lead is 417 degrees Celsius.
Ql = ml * cl * (Tf - Ti)
Now, the total heat gained by the system is equal to zero since no heat is lost to the surroundings.
Qc + Qw + Qi + Ql = 0
Substituting the respective formulas and values:
mc * cc * (Tf - Ti) + mw * cw * (Tf - Ti) + mi * Lf + ml * cl * (Tf - Ti) = 0
Simplifying the equation:
(mc * cc + mw * cw + ml * cl) * (Tf - Ti) = - (mi * Lf)
Now, we can solve for Tf:
Tf = (mi * Lf) / (mc * cc + mw * cw + ml * cl) + Ti
Substituting the given values:
Tf = (0.014 kg * 334000 J/kg + 0.202 kg * 387 J/kg*K * Tf + 0.185 kg * 4186 J/kg*K * Tf + 0.900 kg * 128 J/kg*K * Tf) / (0.202 kg * 387 J/kg*K + 0.185 kg * 4186 J/kg*K + 0.900 kg * 128 J/kg*K) + 0 degrees Celsius
To calculate the final temperature, we need to solve this equation numerically. The final temperature of the system can be obtained through a numerical calculation.