In a physics lab experiment, a student immersed 205 one-cent coins (each having a mass of 3.00 g *.003kg) in boiling water. After they reached thermal equilibrium, she quickly fished them out and dropped them into 0.244 kg of water at 20.0 C in an insulated container of negligible mass.
What was the final temperature of the coins? [One-cent coins are made of a metal alloy - mostly zinc - with a specific heat capacity of 390 J/(kg*K).]
I know that:
Mass,coins=.615 kg
Mass,water=.244 kg
c,coin=390 J/(kg*K)
c,water=418.6 J/(kg*K)
Ti, coins= 100C, 373.15K
Ti, water= 20C, 293.15K
The problem asks for the Tf of the coins, but I'm still missing the Tf of the water...I'm thinking it has something to do with the equilibrium point, but I dont' know that that is!
Heat lost by the coins + heat gained by the water = 0
mass x specific heat x (Tf - Ti) + mass x specific heat x (Tf - Ti) = 0
answered in a duplicate post above.
To solve this problem, you can use the principle of conservation of energy and the equation for heat transfer.
The equation for heat transfer is given by:
Q = mcΔT
Where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, the coins lose heat while the water gains heat. So, the heat lost by the coins is equal to the heat gained by the water:
m_1 * c_1 * (T_f - T_i1) = m_2 * c_2 * (T_i2 - T_f)
Where m_1 is the mass of the coins, c_1 is the specific heat capacity of the coins, T_f is the final temperature of the coins, T_i1 is the initial temperature of the coins, m_2 is the mass of the water, c_2 is the specific heat capacity of water, and T_i2 is the initial temperature of the water.
Plugging in the values:
205 * 0.003 kg * 390 J/(kg*K) * (T_f - 373.15 K) = 0.244 kg * 418.6 J/(kg*K) * (293.15 K - T_f)
Now, you can solve for T_f, the final temperature of the coins.
205 * 0.003 kg * 390 J/(kg*K) * T_f - 205 * 0.003 kg * 390 J/(kg*K) * 373.15 K = 0.244 kg * 418.6 J/(kg*K) * 293.15 K - 0.244 kg * 418.6 J/(kg*K) * T_f
Multiply and simplify:
0.6177 kg * T_f - 273.795 kg*K = 122.3274 kg * K - 0.101 %J
Rearrange the terms:
0.6177 kg * T_f + 122.3274 kg * K = 0.101 J + 273.795 kg*K
Combine like terms:
0.6177 kg * T_f + 122.3274 kg * K = 0.3741 J + 273.795 kg*K
Now, solve for T_f:
0.6177 kg * T_f = 0.3741 J + 273.795 kg*K - 122.3274 kg * K
0.6177 kg * T_f = 152.4416 kg*K
Divide both sides by 0.6177 kg:
T_f = 247.113 K
So, the final temperature of the coins is approximately 247.113 K.