How Much Ice Is Needed to Cool Water from 25°C to 10°C?

In summary, the problem involves calculating the mass of ice needed to reach a final mixing temperature of 10 °C when 5.0 liters of water at 25 °C is mixed with ice at -18 °C. Assuming no heat exchange with the environment, the specific heats and latent heat of ice and water are given. After 2 hours of attempting the problem, the final answer is 0.760 kg, with some discrepancy from the original poster's calculation.
  • #1
Finland
3
0
Latent heat calculation!

Ok... first timer. I am not a science student, done some MBA but wanting some IT course and for entrance exam, I am preparing. I have problem with this one problem. Just provide me a hint. I do calculation part.

'Juice and water is produced from 5.0 liters of water, a temperature of 25 ° C. Ice is added of which temperature is -18 oC. How large ice cube mass should be in order to drink, the final mixing temperature is 10 ° C? Suppose
to drink
and the environment, there is no heat exchange. Juice and water
specific heat is 4.19 kJ / (kg ° C), ice specific heat 2.2 kJ / (kg ° C) and ice
latent heat is 333 kJ / kg. Juice and water density is 1.0 kg/dm3.'

(i just translated this question from finnish to english, so grammar mistake is there!)

Please do provide me hints as to how I proceed with this question.
 
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  • #2


Welcome to Physics Forums.

What are your thoughts on the problem? What have you attempted thus far?
 
  • #3


Hootenanny said:
Welcome to Physics Forums.

What are your thoughts on the problem? What have you attempted thus far?

ha ha 2 hours i attempted this morning... didn't care about breakfast... now i just got it right!

Anyway.. kiitos (thanx)

goes this way
(1kg/dm3 x 5 l) x 4.19 kJ/kgC x 15 = 2.2 kJ/kgC x m x 28 + 333 kJ/kg x m

answer is, m = 0.760 kg (but i got somewhere 0.796) ;-)
 
  • #4


Finland said:
ha ha 2 hours i attempted this morning... didn't care about breakfast... now i just got it right!

Anyway.. kiitos (thanx)

goes this way
(1kg/dm3 x 5 l) x 4.19 kJ/kgC x 15 = 2.2 kJ/kgC x m x 28 + 333 kJ/kg x m

answer is, m = 0.760 kg (but i got somewhere 0.796) ;-)
My answer is the same as yours (0.796kg).
 

Related to How Much Ice Is Needed to Cool Water from 25°C to 10°C?

What is latent heat?

Latent heat is the energy required to change the state of a substance from one phase to another without changing its temperature. This can refer to the energy required to melt a solid into a liquid or to evaporate a liquid into a gas.

How is latent heat calculated?

The formula for calculating latent heat is Q = mL, where Q is the amount of heat energy, m is the mass of the substance, and L is the latent heat value for that substance. The specific latent heat value is dependent on the type of substance and the phase change it is undergoing.

What is the difference between latent heat of fusion and latent heat of vaporization?

The latent heat of fusion refers to the energy required to change a substance from a solid to a liquid, while the latent heat of vaporization refers to the energy required to change a substance from a liquid to a gas. These are different values because the energy required for a substance to undergo a phase change depends on the type of substance and the phase change itself.

Why is latent heat important in understanding weather and climate?

Latent heat plays a crucial role in the formation of weather patterns and the overall climate. When water evaporates from the Earth's surface, it absorbs heat energy and cools the surrounding air, causing it to rise and potentially form clouds and precipitation. This process is important in regulating the Earth's temperature and creating weather patterns.

How is latent heat used in everyday life?

Latent heat is used in many everyday activities, such as cooking, laundry, and air conditioning. For example, when water boils and turns into steam, it absorbs heat energy, which is why we often use hot water to cook food or clean clothes. In air conditioning, the process of evaporating water cools the air and helps to regulate the temperature of a room.

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