PHYSICS 102E
posted by Akitsuke .
(0.2kg)(4190J/kg.degree Celsius)(T70 degrees Celsius)+(0.1kg)(390J/kg.degree Celsius)(T20 degrees Celsius)= 0.
how will I equate it to get the final Temperature? I don't understand how did they get the answer of T=67.8 degrees Celsius.
can someone explain it? Pls. I have an Exam tomorrow and I need to understand it. thanks

Just do the algebra to get the final T. Stick the units (degrees C) on later.
838.2*(T70)+ 39(T20)=0
(838.2+39) T = 58764 + 5800
T = 64564/878.2 = 73 C
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