Question

In a calorimeter, the temperature of 100g of water decreased by 10 degrees Celsius when 10g of ice melted. How much heat was absorbed by the ice?

Answer 1

i have 2 methods \[ m . c . \Delta t \] (m,c, delta t of water) or used \[m . L\] (m= mass of ice melt and L is 334 kJ/kg)

Answer 2

m * c * delta t\[\Delta H= mc \Delta T\]
m = 100g of water
c = 4.18J/g C (its always this for water)
delta T = 10 degrees

Sooo, put it into the equation and you get:
\[\Delta H= (100g)\times (4.18 J/g C) \times (10 C)\]
\[\Delta H= 4180 J\] of heat was absorbed by the ice :)

Answer 3

i prefer with the second method coz 4180 J is the heat released from water, the heat is partially absorbed by ice and the excess is lost to the environment

Answer 4

nothing is lost in a calorimeter