a 1.0kg sample of metal with a specific heat of 0.50KJ/Kg C is 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?
Well, let's get cooking! I mean, calculating!
To find the final temperature, we're going to use the equation:
(metal mass)(metal specific heat)(change in metal temperature) = (water mass)(water specific heat)(change in water temperature)
Let's plug in the numbers:
(1.0kg)(0.50KJ/Kg°C)(Tf - 100°C) = (0.05kg)(4.18KJ/Kg°C)(Tf - 20°C)
Now, we can simplify a bit:
0.5(Tf - 100) = 0.209(Tf - 20)
0.5Tf - 50 = 0.209Tf - 4.18
Now, let's solve for Tf:
0.5Tf - 0.209Tf = -4.18 + 50
0.291Tf = 45.82
Tf ≈ 157.75°C
Voila! The final temperature of the metal and the water is approximately 157.75°C. Just be careful not to cook anything besides the metal and water with that heat!
To find the final temperature of the metal and water, we can use the principle of conservation of heat. The heat gained by the water should be equal to the heat lost by the metal.
First, let's calculate the heat gained by the water:
Qwater = mass × specific heat × change in temperature
Given:
mass of water (mwater) = 50.0g = 0.050kg
specific heat of water (Cwater) = 4.18 kJ/kg°C (specific heat of water is approximately 4.18 kJ/kg°C)
initial temperature of water (Tinitial_water) = 20.0°C
Change in temperature of water:
ΔTwater = final temperature of water - initial temperature of water
Let's substitute the values in the formula.
Qwater = mwater × Cwater × ΔTwater
Next, let's calculate the heat lost by the metal:
Qmetal = mass × specific heat × change in temperature
Given:
mass of metal (mmetal) = 1.0kg
specific heat of metal (Cmetal) = 0.50 kJ/kg°C
initial temperature of metal (Tinitial_metal) = 100.0°C
Change in temperature of metal:
ΔTmetal = initial temperature of metal - final temperature of the metal
Let's substitute the values in the formula.
Qmetal = mmetal × Cmetal × ΔTmetal
Since the heat gained by the water is equal to the heat lost by the metal, we have:
Qwater = Qmetal
mwater × Cwater × ΔTwater = mmetal × Cmetal × ΔTmetal
Now, we can solve for the final temperature of the metal and water.
Let's solve the equation step by step:
1. mwater × Cwater × ΔTwater = mmetal × Cmetal × ΔTmetal
2. ΔTwater = (mmetal × Cmetal × ΔTmetal) / (mwater × Cwater)
3. ΔTwater = (1.0kg × 0.50 kJ/kg°C × (100.0°C - Tfinal_metal)) / (0.050kg × 4.18 kJ/kg°C)
4. ΔTwater = (1.0 × 0.50 × (100.0 - Tfinal_metal)) / (0.050 × 4.18)
5. ΔTwater = 10 × (100.0 - Tfinal_metal) / (0.050 × 4.18)
Now, let's calculate ΔTwater and substitute it back into the equation to find Tfinal_metal.
6. ΔTwater = 10 × (100.0 - Tfinal_metal) / 2.09
Now, let's simplify the equation further:
7. ΔTwater = 4.785 - 0.0477 × Tfinal_metal
Now we can solve for Tfinal_metal:
8. Tfinal_metal = (4.785 - ΔTwater) / 0.0477
We need to find the value of ΔTwater. To do that, we'll use the equation:
ΔTwater = (mwater × Cwater × (Tfinal_water - Tinitial_water)) / mmetal × Cmetal
Given:
Cmetal = 0.50KJ/Kg°C
mwater = 50.0g = 0.050kg
Cwater = 4.18 KJ/Kg°C
Tinitial_water = 20.0°C
Tfinal_water = ?
Plug in the values:
ΔTwater = (0.050 × 4.18 × (Tfinal_water - 20.0)) / (1.0 × 0.50)
Now, we can substitute this value back into the equation for Tfinal_metal to solve for Tfinal_water:
Tfinal_metal = (4.785 - (0.050 × 4.18 × (Tfinal_water - 20.0)) / (0.0477)
Simplifying further, we have:
Tfinal_metal = (4.785 - (0.209 × (Tfinal_water - 20.0))) / 0.0477
Unfortunately, this equation is not easily solved analytically. However, we can use numerical methods or trial-and-error to approximate the final temperature of the metal and water.
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.
First, let's calculate the heat lost by the metal:
Qmetal = mass × specific heat × change in temperature
= 1.0 kg × 0.50 kJ/kg·°C × (final temperature - 100.0°C)
Next, let's calculate the heat gained by the water:
Qwater = mass × specific heat × change in temperature
= 50.0 g × 4.18 kJ/kg·°C × (final temperature - 20.0°C)
Since the heat lost by the metal is equal to the heat gained by the water, we can set up the equation:
Qmetal = Qwater
1.0 kg × 0.50 kJ/kg·°C × (final temperature - 100.0°C) = 50.0 g × 4.18 kJ/kg·°C × (final temperature - 20.0°C)
Now, we can solve the equation to find the final temperature.
1.0 kg × 0.50 kJ/kg·°C × final temperature - 1.0 kg × 0.50 kJ/kg·°C × 100.0°C = 50.0 g × 4.18 kJ/kg·°C × final temperature - 50.0 g × 4.18 kJ/kg·°C × 20.0°C
(1.0 kg × 0.50 kJ/kg·°C - 50.0 g × 4.18 kJ/kg·°C) × final temperature = 50.0 g × 4.18 kJ/kg·°C × 20.0°C - 1.0 kg × 0.50 kJ/kg·°C × 100.0°C
(0.50 kJ/°C - 2.09 kJ/°C) × final temperature = 41.8 kJ - 50.0 kJ
(-1.59 kJ/°C) × final temperature = -8.2 kJ
Divide both sides by -1.59 kJ/°C to isolate the final temperature:
final temperature = -8.2 kJ / (-1.59 kJ/°C)
final temperature ≈ 5.15°C
Therefore, the final temperature of the metal and water is approximately 5.15°C.
heat lost by metal + heat gained by water = 0
[mass metal x specific heat metal x (Tfinal-Tinitial)] | [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Plug in the numbers and solve for Tf. That's the only unknown.