A 2.0-kg piece of steel with a temperature of 70°C is submerged in 1.0 kg of water at 15°C. At what temperature does the water/metal system reach equilibrium? The specific heat of steel is 0.107 cal/(g × K)."
The answer is supposed to be 27 degrees C; however, I keep getting 25 degrees C. Could someone check my method?
m(w)c(w)T(iw) + m(s)c(s)T(is) = m(w)c(w)T(f) + m(s)c(s)T(f)
where m(w) is the mass of the water, c(w) is the specific heat of water in cal/(g K), m(s) is the mass of steel, c(s) is the specific heat of steel, T(iw)is the initial temperature of the water, T(is) is the initial temperature of the water, and T(f) is the final temperature of the system.
I don't like your method, but it might get the right answer.
The sum of the heats gained is zero.
Mw*cw*(Tfw-Tiw)+Ms*Cs(Tfs-Tis)=0
1000*1*(Tf-15)+2000*.107*(Tf-70)=0
Tf(1000+2000*.107)=15000+70*.107*2000
Tf=29980/1214=24.6C
Now, if you used the specific heat of mild steel as .122 cal/gC, which it is, you would get slightly higher temp.
I tried using a method similar to yours earlier, but I got around 70 to 80 degrees C when using it, so I tried another method. I get the same answer of 24.6C when using my method, so I'm confused as to why you don't like it.
Also, when I use the specific heat of mild steel as .122 cal/gC, I get 25.79 C, which is closer to 27 degrees C.
Ah, a hot piece of steel and a cool pool of water, sounds like a daring balancing act! Let's see if we can find the temperature at which they reach equilibrium.
To begin, let's convert the mass of the steel to grams, so we have 2000 grams of steel. Now, let's calculate the heat lost by the steel and the heat gained by the water to achieve equilibrium.
The heat lost by the steel can be calculated using the formula:
Qlost = m × c × ΔT
where m is the mass of the steel, c is the specific heat of steel, and ΔT is the change in temperature.
Let's assume the equilibrium temperature is reached at T°C. So, the temperature change for the steel is ΔT = T - 70°C.
Now, let's calculate the heat gained by the water using the same formula:
Qgained = m × c × ΔT
In this case, the mass of the water is 1000 grams, as given.
Since the heat lost by the steel is equal to the heat gained by the water at equilibrium, we can equate the equations:
m × c × ΔT = m × c × ΔT
Now, we can cancel out the mass and specific heat:
0.107 × (T - 70) = 1.0 × (T - 15)
Let's solve this equation to find the value of T.
0.107T - 7.49 = T - 15
0.107T - T = -15 + 7.49
-0.893T = -7.51
T = -7.51 / -0.893
T ≈ 8.4
So, it seems that our water/metal system will reach equilibrium at approximately 8.4°C. Just a bit cooler than room temperature, but hey, at least they'll be in balance and ready for a refreshing swim together!
To find the temperature at which the water/metal system reaches equilibrium, we can use the principle of heat exchange. The heat gained by the water must be equal to the heat lost by the steel.
First, let's calculate the initial heat content of the water and the steel:
For water:
Mass of water = 1.0 kg
Specific heat of water = 1 cal/(g × K)
Initial temperature of water = 15°C
So, the initial heat content (Q) of water is:
Q_water = mass_water × specific heat_water × change in temperature_water
= 1.0 kg × 1 cal/(g × K) × (15°C - 0°C)
= 1.0 kg × 1 cal/(g × K) × 15 K
= 15 cal
For the steel:
Mass of steel = 2.0 kg
Specific heat of steel = 0.107 cal/(g × K)
Initial temperature of steel = 70°C
So, the initial heat content (Q) of steel is:
Q_steel = mass_steel × specific heat_steel × change in temperature_steel
= 2.0 kg × 0.107 cal/(g × K) × (70°C - 0°C)
= 2.0 kg × 0.107 cal/(g × K) × 70 K
= 15.26 cal
Since heat gained by the water must be equal to the heat lost by the steel at equilibrium, we can equate the heat contents:
Q_water = Q_steel
15 cal = 15.26 cal
This implies that the heat transferred to the water is not enough to bring it to the same temperature as the steel. Therefore, we need to find the additional temperature change for the water.
We can use the formula for heat content to find the additional temperature change for the water:
Q_water = mass_water × specific heat_water × additional temperature change_water
Rearranging the formula:
additional temperature change_water = Q_water / (mass_water × specific heat_water)
= 15 cal / (1.0 kg × 1 cal/(g × K))
= 15 K
So, the additional temperature change for the water is 15°C.
To find the equilibrium temperature, we add this additional temperature change to the initial temperature of the water:
Equilibrium temperature = initial temperature_water + additional temperature change_water
= 15°C + 15°C
= 30°C
Therefore, the water/metal system reaches equilibrium at a temperature of 30°C.