A 60.80 gram sample of iron (with a heat capacity of 0.450 J/g◦C) is heated to 100.00 ◦ It is then transferred to a coffee cup calorimeter containing 52.42 g of water (specific heat of 4.184 J/ g◦C) initially at 20.47 ◦C. If the final temperature of the system is 28.78, what was the heat gained by the calorimeter? If the calorimeter had a mass of 25.19 g, what is the heat capacity of the calorimeter?

Step 1 of 2

Given data: given is 60.80g of iron sample

heat capacity= 0.450 J/g C

water= 52.42 g

specific heat = 4.184 J/ g◦C

Find: To find the heat gained by the calorimeter.


we can use the formula:

q calorimeter​=m calorimeter​×C calorimeter​×ΔT calorimeter​

Step 2 of 2

q calorimeter​=m calorimeter​×C calorimeter​×ΔT calorimeter​


  • m calorimeter​ is the mass of the calorimeter,
  • C calorimeter​ is the specific heat capacity of the calorimeter,
  • ΔT calorimeter​ is the change in temperature of the calorimeter.

m calorimeter​=25.19g,

C calorimeter​ is not given directly, but we can assume it’s the same as water (4.184J/g °C), and

ΔTcalorimeter​ can be calculated as T final​−T initial​.

ΔTcalorimeter​=T final​−T initial​

The final temperature (T final​) is 28.78°C and the initial temperature (T initial​) is 20.47°C,

hence, we can calculate ΔTcalorimeter​.

After finding q calorimeter​, we can then calculate the heat capacity of the calorimeter using the formula:

C calorimeter​= q calorimeterΔTcalorimeter​

Now, let’s calculate these values:

ΔTcalorimeter​= 28.78°C−20.47°C

q calorimeter​= 25.19×4.184×(28.78−20.47)

C calorimeter​= q calorimeter​ΔT calorimeter

C calorimeter​ = 25.19×4.184×(28.78−20.47)28.78°C−20.47°C

C calorimeter​=0.6019 Jg∘C


Performing the calculations will give you the values for qcalorimeter​ and Ccalorimeter​.

Final solution

Therefore, the heat gained by the calorimeter is = 0,6019 J/goC


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