What mass of ice at -14°C will be needed to cool 200cm³ of an orange drink(essentially water) from 25°C to 10°C?

Answer 1

Use ΣmCΔT=0

Let
m1=mass of ice required.
m2=mass of water to be cooled = 200 g
T11=-14
T12=10
T21=25
T22=10

Specific heat of water = 1 cal/g/°C

Lf=latent heat of fusion of ice
= 80 cal/g

Heat lost by water:
mCΔT
=200*1*(25-10)
=3000 cal.

Heat gained by ice
=mCΔT+mLf
=m((1)(10-(-14)) + 80)
=m(240+80)
=320m

ΣmCΔT=0
=>
320m-3000=0
solve for m (= mass of ice)

Answer 2

Mass of ice = x
Heat loss by water = heat gained by ice
Mass=density x volume
Mass of water(orange drink) = 0.001kgcm-3 x 200
= 0.2kg
Heat loss by water= 0.2x4200x(25-10)=12600J
Heat gained by ice
Mc(0-(-14))+mL+mc(10-0)
M[(2100x14)+336000+4200x10]
=407400M
12600=407400m
m=0.031Kg

Answer 3

Mass of ice=?
Temp. of ice= -14°C
Initial temp. of water= 25°C
Final temp. of water= 10°C
Volume of juice(essentially water)=200cm³
SHC of water= 4200J/kg x K
SHC of ice= 2100 J/kg x K
Recall: mass= density x volume
Water has a known density of 1gcm-³
So, m= 1x200= 200g=0.2kg

We know that:
Heat gained by ice=heat loss by water

We know that, H=ml , where l= specific latent heat of fusion
For ice: H= m x 336000= 336000m
H=mct, where m is mass of ice, c is the specific heat capacity of water and t is the final temperature
H=m x 4200 x 10= 42000m
Total heat gained by ice= 336000m + 42000m=378000m

Heat gained by water= mct = 0.2 x 4200 x ( 25-10)= 12600joules
We have that:
378000m = 12600
Therefore, m= 12600÷378000= 0.033kg
Thanks 🤗

Answer 4

150g of ice at 0c is mixed with 300g of water at 50c calculate the temperature of the mixture.

Answer 5

3.3

Answer 6

6.7