Thermal energy question: Final temperature of a solution

[Petronius]

Homework Statement

A 0.400 kg sample of aluminum (c= 9.20 x 102J/kg°C) at 95.0 °C is dropped into a 0.550 kg pot of water which is at 18.0 °C. What temperature will the mixture come to?

Relevant Equations

Qlost+Qgained=0
Qlost= -Qgained
(m_1) (c_1) (∆T_1)= -(m_2 )(c_2 )(∆T_2)
(0.400kg)(9.1 ×10^2 J/kg°C)(T_2 -95.0°C)= (-0.550kg)(4.2×10^3 J/kg°C )(T_2 -18.0°C)
364 (T_2 -95.0°C)= -2310(T_2 -18.0°C)
364T_2 -34580= -2310T_2 +41580)
364T_2+2310 T_2= 41580+34580
2674 T_2= 76160
T_2= 28.4816754°C
T_2=28°C

Hello, thank you for taking the time to read this.. I hope I followed the formatting correctly and this post complies with the expectations of the forum. I solved the equation but my answer seems too low. I tried it again and arrived at the same answer. I also had doubts if it was acceptable for me to rearrange the base equation as " Qlost= -Qgained" . I found this format much easier to contextualize.

This is relatively new topic and as such any feedback would be appreciated.

 

[haruspex]

Homework Statement:: A 0.400 kg sample of aluminum (c= 9.20 x 102J/kg°C) at 95.0 °C is dropped into a 0.550 kg pot of water which is at 18.0 °C. What temperature will the mixture come to?
Relevant Equations:: Qlost+Qgained=0
Qlost= -Qgained
(m_1) (c_1) (∆T_1)= -(m_2 )(c_2 )(∆T_2)
(0.400kg)(9.1 ×10^2 J/kg°C)(T_2 -95.0°C)= (-0.550kg)(4.2×10^3 J/kg°C )(T_2 -18.0°C)
364 (T_2 -95.0°C)= -2310(T_2 -18.0°C)
364T_2 -34580= -2310T_2 +41580)
364T_2+2310 T_2= 41580+34580
2674 T_2= 76160
T_2= 28.4816754°C
T_2=28°C

Hello, thank you for taking the time to read this.. I hope I followed the formatting correctly and this post complies with the expectations of the forum. I solved the equation but my answer seems too low. I tried it again and arrived at the same answer. I also had doubts if it was acceptable for me to rearrange the base equation as " Qlost= -Qgained" . I found this format much easier to contextualize.

This is relatively new topic and as such any feedback would be appreciated.

As you can see, the coefficient for water is nearly five times that for Al, and there is a bit more water than Al, so you would expect the final temperature to be five or six times as far from the initial temperature of the Al as it is from that of the water.
I get very slightly more, though. Plugging in numbers from the start can accumulate rounding errors. Better to work algebraically, arriving at the weighted average
$\theta_f=\frac{m_1c_1\theta_1+m_2c_2\theta_2}{m_1c_1+c_2m_2}$
You can then do all that in one sequence on your calculator.

But just noticed the 9.2 in the statement became 9.1.

 

[Petronius]

Ugh.. I thank you for catching that change to 9.1. I'll redo the calculation following the equation format you provided. Thanks for the input!

 

[haruspex]

Ugh.. I thank you for catching that change to 9.1. I'll redo the calculation following the equation format you provided. Thanks for the input!

Ok, but please do the algebra to obtain that formula for yourself.

 

[Petronius]

As you can see, the coefficient for water is nearly five times that for Al, and there is a bit more water than Al, so you would expect the final temperature to be five or six times as far from the initial temperature of the Al as it is from that of the water.
I get very slightly more, though. Plugging in numbers from the start can accumulate rounding errors. Better to work algebraically, arriving at the weighted average
$\theta_f=\frac{m_1c_1\theta_1+m_2c_2\theta_2}{m_1c_1+c_2m_2}$
You can then do all that in one sequence on your calculator.

But just noticed the 9.2 in the statement became 9.1.

Hello, and thanks again. I followed that equation and calculated it out to 28.5810306. I was unable to punch it all into my calculator at once unfortunately because I did not know how to define the T and we never spent any time covering that in our class. I'm hoping to to find a good tutorial online as that would speed up such calculations.

 

[Lnewqban]

I believe the final temperature value that you have calculated is correct.
If you add that number to the following equation, you will find that the heat released by the aluminum block equals 24442 Joules, which is the same amount of heat absorbed by the water.

$(C_{aluminum})~(m_{aluminum})~(T_{initial~aluminum}-T_{final~system})~=~-(C_{water})~(m_{water})~(T_{initial~water}-T_{final~system})$

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html

 

[Petronius]

I believe the final temperature value that you have calculated is correct.
If you add that number to the following equation, you will find that the heat released by the aluminum block equals 24442 Joules, which is the same amount of heat absorbed by the water.

$(C_{aluminum})~(m_{aluminum})~(T_{initial~aluminum}-T_{final~system})~=~-(C_{water})~(m_{water})~(T_{initial~water}-T_{final~system})$

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html

Thank you for the formula and the link.
I plugged in the values to both sides of the equation that provided and arrived at 24,442.18074 on both sides. This seems like a great way to check my work when I'm dealing with such questions.