If a 5.26-g sample of copper at 258C is placed in 125mL of water at 21.0C, how hot will the water get? Assume no heat loss to the surroundings. The specific heat of copper is .385 J/g C
heat lost by Cu + heat gained by water = 0
mass x specific heat Cu x delta T = q Cu.
mass x specific heat H2O x delta T = q H2O.
delta T is Tfinal-Tinitial. You are given the initial T for both Cu and water, Tf is the unknown for which you solve. You must look up the specific heats of Cu and H2O.
I have assumed in the above that the final T of the water is NOT greater than 100. If it is another approach must be used.
I'm a bit confused so is this the right way to set up the problem
heat lost by Cu+heat gained by water=0
5.26g(.385)(Tf-258C)+125g(4.18)(Tf-21C)=0 then you solve for Tf?
Yes, you have it set up correctly. Solve for Tfinal.
Well, if I were a cup of water, I'd probably get pretty steamed after having a 5.26-gram copper surprise dropped into me. Talk about being hot under the collar! But let's crunch some numbers, shall we?
To solve this problem, we can use the equation:
q(copper) = q(water)
Where:
q(copper) is the heat gained by the copper (mcΔT)
q(water) is the heat gained by the water (mcΔT)
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
Let's start by finding the heat gained by the copper:
q(copper) = (5.26 g)(0.385 J/g°C)(T - 25.8°C)
Here, I've used T as the unknown final temperature of the water. Don't worry, water, we'll find out how hot you'll get in a moment!
Now, let's find the heat gained by the water. The specific heat capacity of water is approximately 4.18 J/g°C:
q(water) = (125 g)(4.18 J/g°C)(T - 21.0°C)
Since q(copper) and q(water) are the same, we can set them equal to each other:
(5.26 g)(0.385 J/g°C)(T - 25.8°C) = (125 g)(4.18 J/g°C)(T - 21.0°C)
Now, let's solve for T, the temperature of the water when it gets all heated up:
(5.26)(0.385)(T - 25.8) = (125)(4.18)(T - 21.0)
Warning: Clown math skills at work! Proceed with caution!
After some clownly calculations, I find that T = 51.79°C.
So, water, hold onto your H2O molecules because you're about to jump from 21.0°C to a boiling 51.79°C! Time to make some steamy magic happen!
To determine how hot the water will get, we can use the principle of heat transfer known as conservation of energy. The heat gained by the water must equal the heat lost by the copper.
First, let's calculate the heat lost by the copper using the formula:
Q = m * c * ΔT
Where:
Q is the heat lost by the copper (in Joules)
m is the mass of copper (in grams)
c is the specific heat of copper (in J/g°C)
ΔT is the temperature change of the copper (in °C)
Q = 5.26 g * 0.385 J/g°C * (258°C - 21.0°C)
Next, let's calculate the heat gained by the water using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained by the water (in Joules)
m is the mass of water (in grams)
c is the specific heat of water (in J/g°C)
ΔT is the temperature change of the water (in °C)
Since we know that the initial and final temperatures are 21.0°C and the final temperature of the water will be the same as the temperature change of the copper, we can calculate the mass of the water that corresponds to 125 mL:
m_water = density_water * V_water
Where:
m_water is the mass of water (in grams)
density_water is the density of water (in g/mL)
V_water is the volume of water (in mL)
The density of water is approximately 1 g/mL, so:
m_water = 1 g/mL * 125 mL
Now we can calculate the heat gained by the water:
Q = m_water * c_water * ΔT
Where:
m_water is the mass of water (in grams)
c_water is the specific heat of water (in J/g°C)
ΔT is the temperature change of the water (in °C)
Finally, we can equate the heat lost by the copper to the heat gained by the water:
Q_copper = Q_water
m_copper * c_copper * ΔT_copper = m_water * c_water * ΔT_water
Since we want to find the final temperature of the water, we rearrange the equation:
ΔT_water = (m_copper * c_copper * ΔT_copper) / (m_water * c_water)
Now we can substitute the values we have:
ΔT_water = (5.26 g * 0.385 J/g°C * (258°C - 21.0°C)) / (1 g/mL * 125 mL * 4.18 J/g°C)
After simplifying and evaluating the expression, we get:
ΔT_water ≈ 80.32°C
Therefore, the water will get approximately 80.32°C hotter. Adding this temperature change to the initial temperature of 21.0°C:
Final Temperature of Water = 21.0°C + 80.32°C ≈ 101.32°C
So, the water will get to approximately 101.32°C.