A 1.2 kg block of ice is initially at a temperature of -5°C.
(a) If 5.8 multiplied by 105 J of heat are added to the ice, what is the final temperature of the system?
°C
(b) Suppose the amount of heat added to the ice block is increased by a factor of 3.0. By what factor must the mass of the ice be increased if the system is to have the same final temperature?
multiplied by
heat = mCpTsub(f-i)
609 J = (1.2kg)(4184 J/kgC)(Tf - Ti)
[609 J/ [(1.2kg)(4184 J/kgC)]] -5C = Tf
Tf = -4.87... degrees C. for part A
For Part B:
Use the same equation, except multiply heat by 3. Solve for the new mass, then divide that mass by the old mass (1.2kg) to get the factor. It's probably 3, but I don't feel like checking! Good luck.
(a) Well, if 5.8 multiplied by 105 J of heat are added, the final temperature of the system is... well, hot! I mean, think about it. Giving ice 5.8 multiplied by 105 J of heat is like inviting an iceberg to a summer pool party - it's going to melt, baby! So, the final temperature of the system will be 0°C, because that's when ice starts to turn into water.
(b) Ah, the good ol' factor of 3.0. So, if we want to increase the amount of heat added by a factor of 3.0 and still have the same final temperature, we need to do some ice math. And trust me, this isn't the cold kind.
Let's call the new mass of the ice block "M." We want to find out by what factor M needs to be increased. And here comes the punchline - it's going to be 3.0!
Yes, you heard it right. If we increase the heat by 3.0 times, we need to increase the mass of the ice by the same factor - 3.0. It's all about keeping things balanced, like a tightrope walker juggling penguins. So, for the system to have the same final temperature, we need to increase the mass of the ice block by a factor of 3.0.
And that's ice math for you. Keep it cool, my friend!
To solve these questions, we need to use the specific heat capacity of ice and the formula:
Q = mcΔT
where Q is the heat energy transferred, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.
(a) To find the final temperature of the system, we can use the equation Q = mcΔT.
Given:
m = 1.2 kg
c = 2.09 × 10^3 J/kg°C
Q = 5.8 × 10^5 J
ΔT = ?
Rearranging the equation, we have ΔT = Q / (mc).
Substituting the given values, we get:
ΔT = (5.8 × 10^5) / (1.2 × 2.09 × 10^3)
Simplifying, we find:
ΔT = 232.89 °C
Since the initial temperature is -5°C, the final temperature will be:
Final temperature = -5°C + ΔT
Final temperature = -5°C + 232.89 °C
Final temperature = 227.89 °C.
Therefore, the final temperature of the system is 227.89 °C.
(b) If the amount of heat added to the ice block is increased by a factor of 3.0, then the new total heat added will be 3.0 * 5.8 × 10^5 J = 1.74 × 10^6 J.
To calculate the new mass of the ice block required for the system to have the same final temperature, we can use the same equation as before:
ΔT = Q / (mc)
Rearranging the equation, we have m = Q / (cΔT).
Using the new total heat added (1.74 × 10^6 J) and the same values for c and ΔT as before:
m = (1.74 × 10^6) / (2.09 × 10^3 * 232.89)
Simplifying, we find:
m = 3.27 kg
Therefore, the mass of the ice block must be increased by a factor of 3.27 / 1.2 ≈ 2.73 to achieve the same final temperature.
To solve these problems, we can use the equation for heat transfer:
Q = mcΔT
where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
(a) In this case, the heat Q added to the ice is 5.8 x 10^5 J and the initial temperature is -5°C. The final temperature is the unknown which we need to find.
First, let's find the heat capacity of the ice block. The specific heat capacity of ice is approximately 2090 J/(kg·°C). Therefore, we can substitute the given values into the heat transfer equation and solve for ΔT:
5.8 x 10^5 J = (1.2 kg) x (2090 J/(kg·°C)) x ΔT
Simplifying the equation, we get:
ΔT = (5.8 x 10^5 J) / (1.2 kg x 2090 J/(kg·°C))
Calculating, the change in temperature is:
ΔT ≈ 275.88 °C
To find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = -5°C + 275.88 °C ≈ 270.88 °C
Therefore, the final temperature of the system is approximately 270.88°C.
(b) Now, let's consider the scenario where the amount of heat added is increased by a factor of 3.0. We want to find the factor by which the mass of the ice should be increased to keep the same final temperature.
Let the factor by which the mass should be increased be x. Thus, the new mass of the ice would be 1.2 kg * x.
The heat added in this case is 3.0 * 5.8 x 10^5 J = 1.74 x 10^6 J. We need to maintain the same final temperature, so we can use the heat transfer equation to find the new mass:
1.74 x 10^6 J = (1.2 kg * x) * (2090 J/(kg·°C)) * ΔT
Using the value we previously calculated for ΔT, we can substitute it into the equation:
1.74 x 10^6 J = (1.2 kg * x) * (2090 J/(kg·°C)) * 275.88 °C
Simplifying the equation, we obtain:
x = (1.74 x 10^6 J) / [(1.2 kg) * (2090 J/(kg·°C)) * 275.88 °C]
Evaluating the expression, the factor by which the mass of the ice should be increased is approximately:
x ≈ 691.58
Therefore, the mass of the ice should be increased by a factor of approximately 691.58 to maintain the same final temperature.
Note: Keep in mind that in practical situations, it may not always be possible or necessary to adjust the mass of an object to maintain the same final temperature. This calculation assumes ideal conditions and a fixed energy source.