In an experiment 50 g of a metal was heated to 100oC and placed in 200 g of a liquid at 25oC. Which of the following combinations of metal and liquid will produce the largest temperature increase in the liquid? (Assume that there is no energy transfer to the surroundings.)
Choose one answer.
a. Al, C2H5OH
b. Au, H2O
c. Cu, CCl2F2
d. Fe, CCl4
Lets look at the math.
Sum of heats gained is zero.
50*Cmetal(Tf-100)+200*Cliquid(Tf-25)=0
Tf(50Cmetal+200Cliquid)=200Cliquid*25+5000Cmetal
Tf=(200Cliquid+5000Cmetal)/(50Cmetal+200Cliquid)
multiplying the right side numerator and deoominator by 1/cliquid
Tf=(200+5000cm/cl)/(50cm/cl+200)
let cm/cl be x.
well, it is apparent to me that the cm/cl term in the numerator predominates, so it needs to be maximized .
So the metal with the largest specific heat, and the liquid with the least.
so on the four choices, compute cm/cl . the winner is the highest ratio.
Well, I must say, we've got quite the cooking show going on here. Let's get started, shall we?
If we want the largest temperature increase in the liquid, we need to find a combination that has a metal with a high specific heat capacity and a liquid with a low specific heat capacity.
Now, let's take a look at our options:
a. Al, C2H5OH: Al, also known as Aluminu-ME, has a relatively low specific heat capacity, and C2H5OH, or ethyl alcohol, has a moderate specific heat capacity. So, this combination may not be the best for our temperature increase.
b. Au, H2O: Au, or gold, has a low specific heat capacity, and water, also known as H2O, has a relatively high specific heat capacity. This combination might just make a golden temperature rise!
c. Cu, CCl2F2: Cu, which stands for copper, has a moderate specific heat capacity, and CCl2F2, or carbon tetrafluoride, has a relatively high specific heat capacity. This combination might not give us the biggest temperature increase.
d. Fe, CCl4: Fe, or iron, has a high specific heat capacity, and CCl4, or carbon tetrachloride, also has a high specific heat capacity. This combination might give us a fair temperature increase, but not the largest.
So, after carefully weighing our options, it seems like the answer is b. Au, H2O. Gold and water, a match made in temperature-increasing heaven!
To determine which combination of metal and liquid will produce the largest temperature increase in the liquid, we need to calculate the amount of heat transferred.
We can use the formula:
q = mcΔT
where:
q = heat transferred
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
Let's calculate the heat transferred for each combination:
a. Al, C2H5OH
The specific heat capacity of aluminum (Al) is 0.897 J/g°C.
The specific heat capacity of ethyl alcohol (C2H5OH) is 2.44 J/g°C.
For Al: q1 = (50 g) * (0.897 J/g°C) * (100°C - 25°C) = 3225 J
For C2H5OH: q2 = (200 g) * (2.44 J/g°C) * (100°C - 25°C) = 34400 J
Total heat transferred: q1 + q2 = 37625 J
b. Au, H2O
The specific heat capacity of gold (Au) is 0.129 J/g°C.
The specific heat capacity of water (H2O) is 4.18 J/g°C.
For Au: q1 = (50 g) * (0.129 J/g°C) * (100°C - 25°C) = 322.5 J
For H2O: q2 = (200 g) * (4.18 J/g°C) * (100°C - 25°C) = 62760 J
Total heat transferred: q1 + q2 = 63082.5 J
c. Cu, CCl2F2
The specific heat capacity of copper (Cu) is 0.386 J/g°C.
The specific heat capacity of CCl2F2 is 0.79 J/g°C.
For Cu: q1 = (50 g) * (0.386 J/g°C) * (100°C - 25°C) = 772.0 J
For CCl2F2: q2 = (200 g) * (0.79 J/g°C) * (100°C - 25°C) = 23660 J
Total heat transferred: q1 + q2 = 24432.0 J
d. Fe, CCl4
The specific heat capacity of iron (Fe) is 0.449 J/g°C.
The specific heat capacity of CCl4 is 0.861 J/g°C.
For Fe: q1 = (50 g) * (0.449 J/g°C) * (100°C - 25°C) = 957.75 J
For CCl4: q2 = (200 g) * (0.861 J/g°C) * (100°C - 25°C) = 30170 J
Total heat transferred: q1 + q2 = 31127.75 J
From the calculations, we can see that option b. Au, H2O will produce the largest temperature increase in the liquid, with a total heat transferred of 63082.5 J. Therefore, the correct answer is b. Au, H2O.
To determine which combination of metal and liquid will produce the largest temperature increase in the liquid, we need to calculate the heat transfer (Q) between the metal and the liquid.
The heat transfer equation is given by Q = mcΔT, where:
- Q is the heat transfer (in joules)
- m is the mass of the substance (in kg)
- c is the specific heat capacity of the substance (in J/kg°C)
- ΔT is the change in temperature (in °C)
Let's go through each combination and calculate the heat transfer for each one:
a. Al, C2H5OH:
- Mass of metal (m1) = 50 g = 0.05 kg
- Mass of liquid (m2) = 200 g = 0.2 kg
- Initial temperature of metal (T1) = 100°C
- Initial temperature of liquid (T2) = 25°C
First, calculate the heat transfer from the metal to the liquid (Q1):
Q1 = m1 * c1 * (T2 - T1)
Q1 = 0.05 kg * (specific heat capacity of Al) * (25°C - 100°C)
b. Au, H2O:
- Repeat the same steps as above, but now using the specific heat capacity of gold (Au) and the mass of water (H2O).
c. Cu, CCl2F2:
- Repeat the same steps as above, but now using the specific heat capacity of copper (Cu) and the mass of CCl2F2.
d. Fe, CCl4:
- Repeat the same steps as above, but now using the specific heat capacity of iron (Fe) and the mass of CCl4.
After calculating Q1 for each combination, compare the values to determine which combination produced the largest temperature increase in the liquid. The combination with the highest Q1 value will produce the largest temperature increase in the liquid.
Note: In the absence of information regarding the specific heat capacities of the substances, it is not possible to provide a definite answer.