The Student Room Group

Physics- Thermal physics

A plastic beaker containing 0.080 kg of water at 15 C was placed in a refrigerator and cooled to 0 C in 1200 Second.
a). Calculate how much energy each second was removed from the water in this process.
The specific heat capacity of water=4200
b). Calculate how long the refrigerator would take to freeze the water in a).
The specific latent heat of fusion of ice= 3.36*10^5
Ans a)== 4.2 J s^-1
b)== 6400 s
Already done part a)(m*c*(T2-T1))/t

But don’t know how to do pat b) can anybody help me please..
Reply 1
Avatar for Kane987
Kane987
7
For part 'b)' you need to use the equation Q = ml (Energy = Mass x Specific Latent Heat)

So you do Q = (0.080) x (3.36x10^5) = 26880 Joules of energy needed

In part 'a)' you worked out that 4.2 Joules of energy was removed per second, so you do 26880/4.2 = 6400s :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you