How much water remains unfrozen after 62.2 kJ is transferred as heat from 364 g of liquid water initially at its freezing point?
1 answer:
You might be interested in
m = mass of the ice added = ?
M = mass of water = 1.90 kg
= specific heat of the water = 4186 J/(kg ⁰C)
= specific heat of the ice = 2000 J/(kg ⁰C)
= latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg
= initial temperature of ice = 0 ⁰C
= initial temperature of water = 79 ⁰C
T = final equilibrium temperature = 8 ⁰C
using conservation of heat
Heat gained by ice = Heat lost by water
m (T -
) + m
= M
(
- T)
inserting the values
m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)
m = 1.53 kg
As per above given data
initial velocity = 19.3 km/s
final velocity = - 18.8 km/s
now in order to find the change in velocity
Part b)
Now we need to find acceleration
acceleration is given by formula
given that
now the acceleration is given as
so above is the acceleration