Asked by bon
When 80.0 grams of a certain metal at 90.0 °C was mixed with 100.0 grams of water at 30.0 °C, the final equilibrium temperature of the mixture was 36.0 °C. What is the specific heat (J/g°C) of the metal?
I suppose I would use this formula:
q= mcÄt
m= mass(g)
q= enthalpy change (j)
c= specifice heat capacity(J/g°C)
Ät= temperature change(°C)
not quite sure how to approach this question!?!?!?!
The sum of the heats added is zero (one is lost, so it is negative).
Heataddedwater + heataddedmetal=0
mw*cw*(Tf-Tiw) + mm*Cm*(Tf-Tim)=0
solve for Cm, the specific heat of the metal.
I suppose I would use this formula:
q= mcÄt
m= mass(g)
q= enthalpy change (j)
c= specifice heat capacity(J/g°C)
Ät= temperature change(°C)
not quite sure how to approach this question!?!?!?!
The sum of the heats added is zero (one is lost, so it is negative).
Heataddedwater + heataddedmetal=0
mw*cw*(Tf-Tiw) + mm*Cm*(Tf-Tim)=0
solve for Cm, the specific heat of the metal.
Answers
Answered by
jacqueline
0.581J/g c
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.