You submerge 100 grams of steel in 200 grams of water. If the steel has an initial temperature of 80°C and the water has an initial temperature of 10°C, what is the final temperature of the system? The specific heat capacity of the steel is 0.15 cal/g•°C
Q=cmΔT
Q1(steel)=Q2(water)
c1•m1• (t1-t) = c2•m2• (t-t2)
628•0.1•(80-t) = 4180•0.2• (t-10)
5024-62.8t = 836t -8360
5024+8360=(836+62.8)t
t=14.9°C
Well, 80°C steel and 10°C water? Things are heating up! Not sure if they'll turn into the dynamic duo of Steamman and Piping Hot, but let's calculate the final temperature just to be safe.
To find the final temperature, we can use the principle of conservation of energy. The total heat gained by the steel should equal the total heat lost by the water.
Heat gained by steel = Heat lost by water
The heat gained by the steel can be calculated using the formula:
Heat gained = mass of steel × specific heat capacity of steel × change in temperature
Heat lost by the water can be calculated using the formula:
Heat lost = mass of water × specific heat capacity of water × change in temperature
Since the final temperature will be the same for the steel and water mix, we can set the two equations equal to each other:
mass of steel × specific heat capacity of steel × change in temperature = mass of water × specific heat capacity of water × change in temperature
Now, let's plug in the values:
100g × 0.15 cal/g•°C × (Tfinal - 80°C) = 200g × 1 cal/g•°C × (Tfinal - 10°C)
Time for some math magic!
0.15(Tfinal - 80) = 200(Tfinal - 10)
Now, let's distribute and simplify:
0.15Tfinal - 12 - 0.15×80 = 200Tfinal - 2000
0.15Tfinal - 12 - 12 = 200Tfinal - 2000
0.15Tfinal - 24 = 200Tfinal - 2000
Now, let's bring all the Tfinal terms to one side:
0.15Tfinal - 200Tfinal = - 24 + 2000
-199.85Tfinal = 1976
Tfinal = 1976 / -199.85
And our final calculation gives us:
Tfinal ≈ -9.88°C
Huh, that seems a bit chilly! It looks like our system got a bit too cold for comfort. Either we found a new way to freeze steel or there might have been an error in our calculations. So, for safety's sake, let's double-check our math or bring a warm coat when dealing with low-temperature steel and water encounters!
To find the final temperature of the system, we can use the principle of heat transfer:
Qsteel = Qwater
The equation for heat transfer is:
Q = mcΔT
Where:
- Q is the heat transferred (in calories)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in cal/g•°C)
- ΔT is the change in temperature (in °C)
First, let's calculate the heat transferred for the steel:
Qsteel = msteel * csteel * ΔTsteel
Substituting the given values:
Qsteel = 100g * 0.15 cal/g•°C * (Final temperature - 80°C)
Now, let's calculate the heat transferred for the water:
Qwater = mwater * cwater * ΔTwater
Substituting the given values:
Qwater = 200g * 1 cal/g•°C * (Final temperature - 10°C)
Since Qsteel and Qwater should be equal, we can set them equal to each other:
Qsteel = Qwater
100g * 0.15 cal/g•°C * (Final temperature - 80°C) = 200g * 1 cal/g•°C * (Final temperature - 10°C)
Simplifying the equation:
15(Final temperature - 80) = 200(Final temperature - 10)
Expanding:
15Final temperature - 1200 = 200Final temperature - 2000
Rearranging:
185Final temperature = 800
Finally, solving for the Final temperature:
Final temperature = 800 / 185 ≈ 4.32°C
Therefore, the final temperature of the system is approximately 4.32°C.
To find the final temperature of the system, we need to consider the heat transfer between the steel and water until they reach thermal equilibrium.
First, let's calculate the heat lost by the steel and the heat gained by the water:
Heat lost by the steel = mass of steel * specific heat capacity of steel * change in temperature of steel
Heat lost by the steel = 100g * 0.15 cal/g•°C * (80°C - final temperature)
Heat gained by the water = mass of water * specific heat capacity of water * change in temperature of water
Heat gained by the water = 200g * 1 cal/g•°C * (final temperature - 10°C)
Since the heat lost by the steel is equal to the heat gained by the water (assuming no heat loss to the surroundings), we can set up the equation:
100g * 0.15 cal/g•°C * (80°C - final temperature) = 200g * 1 cal/g•°C * (final temperature - 10°C)
Now, let's solve for the final temperature:
100g * 0.15 cal/g•°C * (80°C - final temperature) = 200g * 1 cal/g•°C * (final temperature - 10°C)
15(80 - final temperature) = 1(final temperature - 10)
1200 - 15(final temperature) = final temperature - 10
1200 + 10 = 16(final temperature)
1210 = 16(final temperature)
final temperature ≈ 75.6°C
Therefore, the final temperature of the system would be approximately 75.6°C.