A 100. g chunk of pure iron (specific heat capacity 0.444 J/°C*g) with an initial temperature of 100°C was

Question

A 100. g chunk of pure iron (specific heat capacity 0.444 J/°C*g) with an initial temperature of 100°C was placed in a calorimeter containing 135 g of water (specific heat capacity 4.184 J/°C*g) with an initial temperature of 25°C. The final temperature of the calorimeter was 30°C. From this information, calculate the heat capacity (C) of the calorimeter. The correct answer is 56.8 J/°C. Can someone please show me how to get to this answer? Please show steps.

Answer 1

The key element here is to realize that all the heat given off by the iron is absorbed by the calorimeter but not all of it goes into the water (some is lost to the calorimeter itself which is actually what we are trying to find).

So Q_iron = Q_water + Q_calorimeter

Another key thing is that the final temperature of the iron will be the same as the final temperature of the water.

The standard equation for heat capacity is Q = mCΔT so for the water Q = 135*4.184*(30-25) = 2824.2

For the Iron Q = 100*0.444*(30-100) = - 3108 but we can drop the sign because the heat given off is the same as the heat taken in by the water and calorimeter

So 3108 = 2824.2 + Q_calorimeter

Q_calorimeter = 3108 - 2824.2 = 283.8j

But the calorimeter went up in temp by 5 degrees so to get C in J/degree divide the 283.8j by the 5 degrees C to get 56.76 J/°C and using 3 sig figs it becomes 56.8 J/°C