Question

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.2 kg and contains 12 kg of water. A 2.1 kg piece of the metal initially at a temperature of 185°C is dropped into the water. The container and water initially have a temperature of 10.0°C, and the final temperature of the entire system is 18.0°C.

Answer 1

solution. :-) The container has mass 3.2kg and contains 12 kg water. The piece of metal is 21 kg and initially at temperature of 185c. The temperature change 10.0°c to 18と The Heat transferred from the piece of metal equals the Heat transferred of the container and water Op = Owt ac MP CATp = Me (WAT + MecAT where p,w and a refer to piece of metal, water and container, respectively Thus mw ( WDT C Mp D Tp-MCAT substitute the givens and the known values 12 kg *(4187 TIKS K) (18Ć– loc) C= 2.1 kg x (185°-1807 - 3.219 (186-100)

solution. :-) The container has mass 3.2kg and contains 12 kg water. The piece of metal is 21 kg and initially at temperature of 185c. The temperature change 10.0°c to 18と The Heat transferred from the piece of metal equals the Heat transferred of the container and water Op = Owt ac MP CATp = Me (WAT + MecAT where p,w and a refer to piece of metal, water and container, respectively Thus mw ( WDT C Mp D Tp-MCAT substitute the givens and the known values 12 kg *(4187 TIKS K) (18Ć– loc) C= 2.1 kg x (185°-1807 - 3.219 (186-100)

401952 प 19-6 350 - 7 - 25.6) 401962 325. J K5- C: 1236.39_J kg-k