Question

A 85.0 gram piece of iron at 25 degrees C is put into 635 grams of water at 15.0 degrees C. What is the final temperature of the water ad the iron

Answer 1

Explanation:

Let the final temperature is T

The amount of heat gained by the water,Q1= m1S1(T-T1)

= 0.635×4200×(T-15)

= 2667(T-15)

The amount of heat released by the iron,Q2= m2S2(T2-T)

= 0.065×448×(525-T)

= 29.12(525-T)

According to the principle of thermometry,

Q2=Q1

or,29.12(525-T)=2667(T-15)

or,525-T= 91.59(T-15)

or,525-T=91.59T-1373.85

or, 92.59T=1898.85

or,T= 20.51°C