• CONTEST ALERT - Experience the power of DDR5 memory with Kingston Click for details

Physics Numericals

Status
Not open for further replies.

Chirag

Cyborg Agent
Help me with this sum.. got no clue what to do..

Q) A piece of iron of mass 200g and temperature 300 degree C is dropped into 1.00 kg of water of temperature 20 degree C. What will b e the eventual temperature of the water? ( Take c for iron as 470 J/kg and for water as 4200 J/kg ).
 

ring_wraith

=--=l33t=--=
mass of iron = 0.2 kg
t of iron = 300c
m of water = 1
initial t of water = 20
let final temp be t2

q released by iron = q absorbed by water.
0.2*470*(300-t2) = 1*4200*(t2-20)

Just simplify and obtain value of t2.
 
Status
Not open for further replies.
Top Bottom