A 25.5-g aluminum block is warmed to 65.4 °C and plunged into
an ...

Question

A 25.5-g aluminum block is warmed to 65.4 °C and plunged into
an insulated beaker containing 55.2 g water initially at 22.2 °C.
The aluminum and the water are allowed to come to thermal
equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?

Accepted Answer

hey everyone in this example, we have a 3.51 g silver ball initially placed at 10 degrees Celsius, and a container with 15 11 g of water With a temperature of 64°C. So we want to recognize that the silver balls at the lower temperature and this question is asking us to determine the final temperature when the silver ball and water reach thermal equilibrium. And so thermal equilibrium is the key word here. And that means that we're comparing the two heats between our objects are silver ball and our water. And so we want to recall that the object that has the higher temperature will have a negative value for heat. And because this is an equilibrium, it's going to equal the heat for the silver ball at some point during equilibrium where the silver ball will have a positive value for heat. And that's because it's the object at the lower temperature. So it's going to gain heat from the water that will that will give off heat to the silver ball. So just as a diagram for an example, this is our container of water, This is our silver ball and our silver ball is going to absorb the heat from the water because the water is at the higher temperature and the silver is at the lower temperature. And so now that we have this relationship expressed, we can recall that we can substitute this relationship so that we have a positive value for M cat equal to a negative value for M cat. And so we want to recall that cat stands for the mass times the specific heat for object and then times delta T. Which is the final minus initial temperature. So And um substituting the information from the problem for the mass of the silver ball. Were given a value of 3.51 g. And then for the specific heat of silver we would look that up and see that it's equal to a value of 0.2 35 jewels divided by grams times degrees Celsius. We want to recall that specific heater metals are typically low numbers. So this number should make sense. And then we're multiplying by the Delta T. Which is the difference in temperature. So we don't know the final temperature. That's what we're trying to solve for. But we do know the initial temperature has given us 10°C. So we're going to set this equal to a negative value on the other side for our object at the higher temperature being our water which is given to have a mass of 15.11 g. We're also going to recall the specific heat for water which we recall is 4.184 jewels divided by g times degrees C. And then lastly we want the Delta T. Where we take the final temperature which is what we're trying to solve for minus the initial temperature given us 64°C. So our next step is to simplify this. And in order to simplify this, we're going to multiply these two terms on both sides. And so what we're going to get in our next line when we do so is a value of 0.82485 jules divided by degrees Celsius. And that's because grams will cancel out in this first step And then we're going to have that multiplied by TF -10°C. On the other side, we should have a negative value equal to 63.2202 jules divided by degrees Celsius because grams again will cancel out. And let me just use a different color to make that clear here, which you know is canceling. And so this is going to be multiplied by TF -64°C. And so our next step is to simplify this even further by distributing this term here on both sides to our parentheses terms. So we're distributing these two terms to our parentheses. And so in our calculators, when we do so we're going to get a value of 0.82485 T. F jules divided by degrees Celsius multiply or sorry, subtracted by 8.2485 jewels. And then that is going to equal on the other side. We should get negative 63-2. TF jewels divided by degrees Celsius Added to 4046.9 Jules. And so now we can simplify this even further first by adding on the right hand side. The 63.22 oh two TF jewels divided by degrees Celsius term. And so that's going to cancel out on the right hand side. And we will also make the change of adding 8. jewels to both sides to simplify this even more. And so we're going to cancel this on the left hand side. And so what we would get in our next line Is 63.2202. And sorry, let me actually correct that because that is not the value, it's going to be equal to 64. 451. We still have the TF term attached and then we still have the units of jewels divided by degrees Celsius. And that is equal on the other side to a value of 4054.3 for jules. And so our next step is to isolate for final temperature. So we're going to divide by the term on the left hand side, 64.04 51 jules divided by degrees Celsius or the units that it carries. And so this is going to allow us to cancel out our units of jewels on the right hand side as well as this term here. And so what we would be left with is a value equal to just TF Equal to 63.30°C. And so this will complete this example as our final answer for our final temperature of both the water and the silver ball when thermal equilibrium is reached. So I hope that everything we went through is clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

A 25.5-g aluminum block is warmed to 65.4 °C and plunged into an insulated beaker containing 55.2 g water initially at 22.2 °C. The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?

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