- #1
Petronius
- 13
- 2
- 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.
This is relatively new topic and as such any feedback would be appreciated.