Asked by Dorrito
When 2.1x10^3 J of heat is added to 100.g of a substance, its temperature increases from 15 C to 40 C. What is the specific heat capacity of the substance?
Answer is 8.4x10^2 but I need to see the work involved in how to solve the problem
Answer is 8.4x10^2 but I need to see the work involved in how to solve the problem
Answers
Answered by
drwls
heat added = Qin = M*C*(delta T)
You must have seen that equation before. It is how specific heat is defined. Solving for C,
C = Qin/(M*C) = 2100J/(0.100 kg*25C)
= 840 J/(kg*C)
"delta T" is the increase in temerature.
You must have seen that equation before. It is how specific heat is defined. Solving for C,
C = Qin/(M*C) = 2100J/(0.100 kg*25C)
= 840 J/(kg*C)
"delta T" is the increase in temerature.
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