Chemistry Forum

A 48.59 gram sample of iron (with a heat capacity of 0.450 J/g ºC) is heated to 100.0 ºC. It is then transferred to a coffee cup calorimeter containing 66.51 g of water (specific heat of 4.184 J/ g ºC) initially at 20.63 ºC. If the final temperature of the system is 23.59 ºC, how much heat was absorbed by the calorimeter? Record your answer as a whole number (assume the sign is positive).

I am working on this question. Any chemistry help, hint or correct answer will be accepted.

It is not asked. You can calculate the heat what was absorbed by the water.
But you could calculate the Initial temperature of iron.

This reaction released 1670.743 J of heat out of which 847.04 Jwas absorbed by coffee-cup calorimeter and 823.70 J was absorbed by water.

The absorbtion of 823,7 kJ/ mol for water is correct. But how you get the other values. What is the 7nitial temperature of iron?

Initial temperature of iron was 100.0℃ and its final temperature is equal to final temperature of the system which is 23.59℃. So ∆T for 48.59 g sample of iron is 100.0℃- 23.59℃= 76.41℃.
If you read carefully, you will notice that the questioner asked how much heat absorbed by coffee cup calorimeter, not by water.

True, but calorimeter constant is unknown.