a 1.0 kg sample of metal with a specific heat of 0.50KJ/kgCis heated to 100.0C and then placed in a 50.0g sample of water at 20.0C. What is the final temperature of the metal and the water?
I can't figure this out. Is the answer room temperature???
How did you get 4.184 J in the equation?
No. The sum of heats gained is zero.
heat gained by water + heat gained my metal=0
mw*cw*(Tf-20)+mm*cm*(Tf-100)=0
solve for Tf.
[1.0Kg][1000g][100C-Tf][0.50KJ][ 1Kg][1000J] = [50.0g][Tf-20C][4.184 J]
[ 1Kg] [ KgC][1000g][ 1KJ] [ gC ]
[1000g][100C-Tf][0.50J] = [50.0g][Tf-20C][4.184 J]
[ gC] [ gC ]
[1000g][100C-Tf][0.50 J/gC] = [Tf-20C]
[ 50.0g] [4.184 J/gC]
[2.39][100C-Tf] = [Tf-20C]
239C-2.39Tf = Tf-20C
239C + 20C = Tf + 2.39Tf
259C = 3.39Tf
259C = Tf
3.39
76C = Tf
Well, let me see if I can heat things up and help you out with this question!
To find the final temperature of the metal and the water, we can use the principle of conservation of energy. The heat lost by the metal will be equal to the heat gained by the water.
The heat lost by the metal is given by:
Qmetal = mmetal * cmetal * ΔTmetal
Where mmetal is the mass of the metal (1.0 kg), cmetal is the specific heat of the metal (0.50 kJ/kgC), and ΔTmetal is the change in temperature of the metal.
The heat gained by the water is given by:
Qwater = mwater * cwater * ΔTwater
Where mwater is the mass of water (50.0 g), cwater is the specific heat of water (4.18 kJ/kgC), and ΔTwater is the change in temperature of water.
Since the heat lost by the metal is equal to the heat gained by the water, we can write:
Qmetal = Qwater
mmetal * cmetal * ΔTmetal = mwater * cwater * ΔTwater
Plugging in the given values:
(1.0 kg) * (0.50 kJ/kgC) * ΔTmetal = (0.050 kg) * (4.18 kJ/kgC) * ΔTwater
Simplifying:
0.50 * ΔTmetal = 4.18 * ΔTwater
Now, rearranging to solve for ΔTwater:
ΔTwater = (0.50 * ΔTmetal) / 4.18
Substituting the values given in the problem, with the metal initially heated to 100.0C and the water initially at 20.0C:
ΔTwater = (0.50 * (100.0 - ΔTwater)) / 4.18
Solving the equation gives us ΔTwater ≈ 16.06C.
Therefore, the final temperature of the water is approximately 20.0C + 16.06C = 36.06C.
Finally, to find the final temperature of the metal, we can substitute this value back into the equation:
ΔTwater = 100.0C - ΔTmetal
16.06C = 100.0C - ΔTmetal
Solving for ΔTmetal gives us ΔTmetal ≈ 100.0C - 16.06C ≈ 83.94C.
So, the final temperature of the metal is approximately 83.94C.
And there you have it! The final temperature of the metal and the water are 83.94C and 36.06C, respectively. Not quite room temperature, but hopefully I was able to heat up your understanding of this problem!
To find the final temperature of the metal and the water, we can use the principle of heat transfer. The heat gained by the water will be equal to the heat lost by the metal.
First, let's calculate the heat gained by the water. We can use the equation:
Q = m * c * ΔT
Where:
Q = heat gained
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
Substituting the given values:
m = 50.0 g (converted to kg: 50.0 g ÷ 1000 = 0.050 kg)
c = 4.18 kJ/kg°C (specific heat capacity of water)
ΔT = (final temperature - initial temperature) = (Tf - 20.0°C)
Q_water = 0.050 kg * 4.18 kJ/kg°C * (Tf - 20.0°C)
Next, let's calculate the heat lost by the metal. We use the same equation:
Q = m * c * ΔT
Where:
Q = heat lost
m = mass of metal
c = specific heat capacity of the metal
ΔT = change in temperature
Substituting the given values:
m = 1.0 kg
c = 0.50 kJ/kg°C (specific heat capacity of the metal)
ΔT = (final temperature - initial temperature) = (Tf - 100.0°C)
Q_metal = 1.0 kg * 0.50 kJ/kg°C * (Tf - 100.0°C)
Since the heat gained by the water is equal to the heat lost by the metal (Q_water = Q_metal), we can set up an equation:
0.050 kg * 4.18 kJ/kg°C * (Tf - 20.0°C) = 1.0 kg * 0.50 kJ/kg°C * (Tf - 100.0°C)
Simplifying the equation:
0.209(Tf - 20.0) = 0.5(Tf - 100.0)
Now, we can solve for Tf.
0.209Tf - 4.18 = 0.5Tf - 50.0
0.291Tf = 45.82
Tf ≈ 157.7°C
So, the final temperature of both the metal and the water is approximately 157.7°C, not the room temperature.