Discovering the Initial Temperature of a Heated Metal Bolt with Thermodynamics

[laker88116]

A 0.50 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capcity of 899 J/kg * C, find the initial temparature of the metal.

I am not sure even where to start.

 

[pt176900]

start with the conservation of energy. The initial and final energies of the system should be the same.

 

[laker88116]

i know that much, but i don't know which equation to use

 

[laker88116]

1/2(m)(mu)^2=constant, i figure but how do i find temperature from that

 

[witze]

i know that much, but i don't know which equation to use

You're supposed to be able to derive equations as simple as that, not necessarily remember them.

 

[witze]

1/2(m)(mu)^2=constant, i figure but how do i find temperature from that

What the hell is that?! BTW, this thread should be in the Homework Help section.

The transferred energy equals change in temp. of substance times its mass times its heat capacity (assuming, of course, that the heat capacity is given as energy per unit temp per unit mass). Energy lost or gained by the metal equals energy lost or gained by the water...

 

[laker88116]

or is it cmt of water = cmt of bolt?

 

[The Bob]

A 0.50 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capcity of 899 J/kg * C, find the initial temparature of the metal.

I am not sure even where to start.

Now I am not completely sure as I have only studied this, so far, in chemistry but here goes.

You can work out the Energy Transferred to the water:
Energy Transferred / E = Specific Heat Capacity / c x Mass of Water / m x Change in Temperature / ΔT

Specific Heat Capacity for water is $$4.17 Jg^{-1}K^{-1}$$

$$E = 4.17 Jg^{-1}K^{-1} \times 150g \times (25 - 21)$$

$$E = 625.5 \times 4 = 2502 J$$

From this you can then do the reverse for the bolt.

From here you can probably find something that will help and give an answer or someone will come along and say I am wrong. Either way.

The Bob (2004 ©)

 

[laker88116]

um, well the answer is 81 degrees celsius, if u solve for (25-x) in that equation ur given delta t, which should be change from some reference point in the problem which I am not seeing

 

[The Bob]

The 2502J is the energy that the water transfers to its surroundings. This is the energy that the bolt must have given the water.

The Bob (2004 ©)

 

[laker88116]

Now I am not completely sure as I have only studied this, so far, in chemistry but here goes.

You can work out the Energy Transferred to the water:
Energy Transferred / E = Specific Heat Capacity / c x Mass of Water / m x Change in Temperature / ΔT

Specific Heat Capacity for water is $$4.17 Jg^{-1}K^{-1}$$

$$E = 4.17 Jg^{-1}K^{-1} \times 150g \times (25 - 21)$$

$$E = 625.5 \times 4 = 2502 J$$

From this you can then do the reverse for the bolt.

$$2502 J = 899 Jg^{-1}K^{-1} \times 500g \times (25 - x)$$

From here you can probably find something that will help and give an answer or someone will come along and say I am wrong. Either way.

The Bob (2004 ©)

The problem with your first equation is that the (25-21) is in celsius and the 4.17 j * g^-1 * k^ -1 is in kelvin.

 

[laker88116]

er never mind taht doesn't matter

 

[laker88116]

i don't know i can't get the answer I've tried a lot of things

 

[The Bob]

The problem with your first equation is that the (25-21) is in celsius and the 4.17 j * g^-1 * k^ -1 is in kelvin.

I was going to say. This really doesn't matter so long as the units are the same (and not Fahrenheit ).

The Bob (2004 )

 

[The Bob]

Personally I feel the heat capacity of the bolt is too high but that might just be me.

The Bob (2004 ©)

 

[laker88116]

i agree that's what i was getting at

 

[The Bob]

I think I might be close to cracking it. The number you gave was $$899 J kg^{-1}$$

I needed it in $$Jg^{-1}$$

This means that I need to convert it. This means that $$899 J kg^{-1} = 899000Jg^{-1}$$

The Bob (2004 ©)

 

[The Bob]

The amount of energy needed to increase the temperature of the water by 4°C is going to be what has already been worked out:

$$E = (4.17 Jg^{-1} K^{-1})(150g)(25-21) = 2502J$$

So this is the energy give to the water. This is also the energy give up by the metal bolt. So:

$$2502J = (899 Jkg^{-1}K^{-1})(0.5kg)(T_i - 25)$$ when $$T_i$$ is the inital temperature.

I got most of the help from here. See if you can interprete it.

I think the specific heat capacity is wrong. I think it should be 0.08.

The Bob (2004 ©)

 

[laker88116]

hmm, i see, I am really close to getting it, i think u have to convert the top to kg as well

 

[laker88116]

im not sure but if my memory serves me right energy is hand in hand with kg, not g.

 

[laker88116]

no that can't be right, i think the way u have it is right, its just not getting the answer i have

 

[The Bob]

I think the answer, for your heat capacity, is 26.6°C. I have thought logically about it and thought the steps through but I can't see anything wrong.

The Bob (2004 ©)

 

[The Bob]

Here is my thinking:

1. You need to know how much energy has to be transferred to the water for it to raise 4°C. This energy is, almost without a doubt, 2502J.

2. This is the energy that the metal bolt must give to the water by cooling from an unknown temperature to 21°C. I believe the equation I set up for it is right.

3. Then you just have to solve it.

The Bob (2004 ©)

 

[laker88116]

that makes perfect sense to me, i just don't see why it doesn't yield the right answer

 

[The Bob]

I am stuck on that as well. The metal bolt needs 899 Joules of energy to change the temperate of one kilgram of the metal by a degree (in temperature). This means it doesn't need to be heated too much for there to be a lot of energy that can be transferred by cooling.

Where has the answer come from?

The Bob (2004 ©)

P.S. I have also decided that the temperature shuold be 30.56°C.

 

[laker88116]

my physics teacher, its not his question though so its possible he could be wrong

 

[The Bob]

my physics teacher, its not his question though so its possible he could be wrong

Well I am definitely sticking with 30.56°C unless I can think of any reason why it shouldn't be. As for your teacher, just prove him wrong and go through your (our) logic behind the question and getting the answer.

The Bob (2004 ©)

 

[laker88116]

thanks for the help, i understand it a lot better now

 

[The Bob]

thanks for the help, i understand it a lot better now

That's alright. I have a better understanding as well. My teacher never told me what I was working out just that I had to. Never really looked into it because I ended up writing 39 pages on the enthply change of combustion of alchols using that equation. I used it so much and yet never understood it. Crazy.

I will be interested to know what the answer is though. If we got it right or not. PM me with the right answer and if 30.56°C is wrong then reason why it is.

Cheers

The Bob (2004 ©)