Question

What is the amount of heat required to increase the relative temperature of an object by 0.8cal / gm ° c to 20 c of a mass of 50 g? ​

Answer 1

Answer:

0.43jg

Explanation:

Mass of the metal, m = 0.20 kg = 200 g 

Initial temperature of the metal, T1 = 150 oC

Final temperature of the metal, T2 = 40 oC

Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g

Volume of water, V = 150 cm3

Mass (M) of water at temperature T = 27oC:

150×1=150g

Fall in the temperature of the metal:

ΔTm=T1-T2 =150−40=110oC

Specific heat of water, Cw=4.186J/g/K

Specific heat of the metal =C

Heat lost by the metal,  =mCT .... (i)

Rise in the temperature of the water and calorimeter system: T1−T=40−27=13oC

Heat gained by the water and calorimeter system: =m1CwT=(M+m)CwT ....(ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

mCΔTm=(M+m)CwTw

200×C×110=(150+25)×4.186×13

C=(175×4.186×13)/(110×200)=0.43Jg