Question regarding Calorimetry?

  • #1
139
0

Homework Statement


A copper sample whose mass mc is 75 g is heated in a laboratory oven to a temperature of 312 degrees Celsius. The copper is then dropped into a glass beaker containing mass mw = 220 g of water. The effective heat capacity of the beaker is 190 J/K. The initial temperature of the water and beaker is 12 degrees Centigrade. What's the common final temperature of the copper, glass, and water?


Homework Equations


Q = mcΔT
(mw)(cw)(ΔTw) = (mc)(cc)(ΔTc)



The Attempt at a Solution


I tried substituting the given values into the equation but I was confused about what to do with the heat capacity of the beaker.
 

Answers and Replies

  • #2
gneill
Mentor
20,913
2,861
I tried substituting the given values into the equation but I was confused about what to do with the heat capacity of the beaker.
They don't give you the heat capacity per gram of the beaker or its mass, instead they give you the heat capacity itself. That is, the heat capacity of the beaker as a whole, as if you had a "cb" and "mb" and multiplied them: Cb = cb*mb.

Here's a shortcut you can use when there are no phase changes involved (in this case it's a bit dubious because they don't tell you whether or not any steam generated at first contact of the 312C copper with the water will be contained inside the beaker), Nevertheless...

Convert all the heat-capacities-per-gram to heat capacities for the objects or substances (multiply the given J/kg/K values for the objects by the mass of those objects). Sum them up to find a grand total heat capacity of the system. Next, find the total amount of heat in the system to begin with. That is, sum the heats (Joules) in the individual objects or substances. When steady state is reached the heat will be spread evenly though the whole system. So the total heat and total heat capacity can be used to find the uniform temperature.
 

Related Threads on Question regarding Calorimetry?

  • Last Post
Replies
2
Views
616
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
3K
Replies
1
Views
765
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
3
Views
2K
Top