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?
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 🤗
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
150g of ice at 0c is mixed with 300g of water at 50c calculate the temperature of the mixture.
To answer this question, we need to calculate the heat transfer. The heat transfer can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat transfer (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
First, we need to calculate the heat transfer to cool the orange drink from 25°C to 10°C. Since the orange drink is essentially water, we can use the specific heat capacity of water, which is approximately 4.18 joules per gram per degree Celsius.
Given:
Initial temperature (T1) = 25°C
Final temperature (T2) = 10°C
Volume of orange drink (V) = 200 cm³
To calculate the mass (m) of the orange drink, we need to convert the volume to grams using the density of water, which is approximately 1 g/cm³.
Density = mass/volume
Mass = density * volume
Since the density of water is 1 g/cm³, the mass of the orange drink is:
Mass = 1 g/cm³ * 200 cm³
Mass = 200 grams
Now we can calculate the heat transfer:
Q = m * c * ΔT
Q = 200 g * 4.18 J/g°C * (10°C - 25°C)
Q = 200 g * 4.18 J/g°C * (-15°C)
Q = -12540 J
Note that the negative sign indicates that heat is being lost by the orange drink.
The overall heat transfer can be considered as the heat gained by the ice. The heat gained by the ice can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained by the ice (in joules),
m is the mass of the ice (in grams),
c is the specific heat capacity of ice (in joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
The specific heat capacity of ice is approximately 2.09 joules per gram per degree Celsius.
Since we know the heat transfer (Q) and the specific heat capacity of ice (c), we can rearrange the formula to calculate the mass of the ice (m):
m = Q / (c * ΔT)
m = -12540 J / (2.09 J/g°C * (0°C - (-14°C)))
m = -12540 J / (2.09 J/g°C * (-14°C))
m = -12540 J / (-29.26 J/g)
m ≈ 428.8 g
So, approximately 428.8 grams of ice at -14°C will be needed to cool 200 cm³ of the orange drink from 25°C to 10°C.
3.3
6.7
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)