A 0.300 kg iron horseshoe that is initially at 550°C is dropped into a bucket containing 19.4 kg of water at 21.1°C. What is the final equilibrium temperature (in °C)? Neglect any heat transfer to or from the surroundings. Do not enter units.
I'm taking (specific heat of water*mass of water)+(specific heat of iron*mass of iron) and keep getting the wrong answer. Please help!
To find the final equilibrium temperature, we can use the principle of conservation of energy. The heat lost by the iron horseshoe will be equal to the heat gained by the water in the bucket.
1. Start by calculating the heat lost by the iron horseshoe using the formula:
Q1 = mass of iron * specific heat of iron * change in temperature of iron
Q1 = 0.300 kg * 450 J/kg°C * (550°C - Tf) [where Tf is the final temperature]
2. Next, calculate the heat gained by the water using the formula:
Q2 = mass of water * specific heat of water * change in temperature of water
Q2 = 19.4 kg * 4186 J/kg°C * (Tf - 21.1°C)
3. Since the heat lost by the iron horseshoe is equal to the heat gained by the water, we can set Q1 equal to Q2:
0.300 kg * 450 J/kg°C * (550°C - Tf) = 19.4 kg * 4186 J/kg°C * (Tf - 21.1°C)
4. Simplify the equation by multiplying and distributing:
135 J/°C * (550°C - Tf) = 80942 J/°C * (Tf - 21.1°C)
5. Expand the equation further:
74250 J - 135 J*Tf = 80942 J*Tf - 80942 J*21.1°C
6. Bring all the terms with Tf to one side of the equation and the constant terms to the other side:
80942 J*Tf + 135 J*Tf = 80942 J * 21.1°C + 74250 J
7. Combine like terms and evaluate the equation:
(80942 J + 135 J)*Tf = 80942 J * 21.1°C + 74250 J
(81077 J)*Tf = 80942 J * 21.1°C + 74250 J
Tf = [80942 J * 21.1°C + 74250 J] / 81077 J
8. Calculate the final temperature Tf:
Tf ≈ 29346.62 °C / 81077 J
Tf ≈ 36.2 °C
The final equilibrium temperature (Tf) is approximately 36.2°C.
To find the final equilibrium temperature, you can use the principle of conservation of energy. The heat gained by the water must be equal to the heat lost by the iron horseshoe.
The formula to calculate the heat gained or lost by an object is Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature.
First, let's calculate the heat gained by the water:
Qwater = mwater * cwater * ΔTwater
Given:
mwater = 19.4 kg (mass of water)
cwater = 4186 J/(kg°C) (specific heat of water)
ΔTwater = Tf - 21.1°C (change in temperature of the water, where Tf is the final equilibrium temperature)
Next, let's calculate the heat lost by the iron horseshoe:
Qiron = miron * ciron * ΔTiron
Given:
miron = 0.300 kg (mass of iron horseshoe)
ciron = 450 J/(kg°C) (specific heat of iron)
ΔTiron = Tf - 550°C (change in temperature of the iron, where Tf is the final equilibrium temperature)
Since we know that the heat gained by the water is equal to the heat lost by the iron, we can set up the equation:
Qwater = Qiron
mwater * cwater * ΔTwater = miron * ciron * ΔTiron
Substituting the known values:
(19.4 kg) * (4186 J/(kg°C)) * (Tf - 21.1°C) = (0.300 kg) * (450 J/(kg°C)) * (Tf - 550°C)
Simplifying the equation:
(80644.4 J/°C) * (Tf - 21.1°C) = (135 J/°C) * (Tf - 550°C)
Now, you can solve for Tf by isolating it on one side of the equation:
(80644.4 Tf - 80644.4 * 21.1) = (135 Tf - 135 * 550)
Rearrange the terms:
80644.4 Tf - 1702692.4 = 135 Tf - 74250
Combine like terms:
(80644.4 - 135) Tf = -74250 + 1702692.4
Simplify:
80479.4 Tf = 1628442.4
Finally, solve for Tf:
Tf = 1628442.4 / 80479.4
Tf ≈ 20.24°C
So, the final equilibrium temperature (Tf) is approximately 20.24°C.