Question

Given 355 mLmL of water at 24 ∘C∘C, how much ice (in grams) would have to be added to this water so that after the ice has melted, the overall temperature of the system has been lowered to 5 ∘C∘C? (Assume the density of water is 1.0 g/mLg/mL.)

Answer 1

Answer:

Mass of ice required = 79.4 g

Explanation:

Mass of water = density of water * volume of water

mass of water = 355 mL * 1.0 g/mL = 355 g

Quantity of heat emitted by water during the process of the melting of ice is given as, Q = mcΔT

where c is specific heat capacity of water = 4.18 J/g/°C

ΔT is temperature change = final temperature - initial temperature

m is mass of water

Q = 355 g * 4.18 J/g/°C * (5 - 24)°C

Q = -28194.1 J

Let the mass of ice that would absorb this heat be m

Total heat absorbed by the ice = heat  absorbed in fusion of ice at 0°C to water at 0°c + heat absorbed in changing from water at 0°C to water at 5°C

Heat  absorbed in fusion of ice at 0°C to water at 0°c, H₁ = mL

where m is mass of ice; L is latent heat of fusion of ice = 334 J/g

H₁ = m * 334 J/g

H₁ = (334 J/g)m

Heat absorbed in changing from water at 0°C to water at 5°C, H₂ = mcΔT

H₂ = m * 4.1 J/g/°C * (5 - 0)°C

H₂ = (20.9 J/g)m

Total heat absorbed = (334 J/g)m + (20.9 J/g)m = (354.9 J/g)m

Since heat evolved by water = heat absorbed by ice

28194.1 J = (354.9 J/g)m

m = 28194.1 J/354.9 J/g

m = 79.4 g

Therefore, mass of ice required = 79.4 g

Answer 2

The mass of ice that results to lower the temperature of water from 25$\rm ^\circ C$ to 5$\rm ^\circ C$ has been 79.4 grams.

From the given volume the mass of water can be calculated as:

mass = density $\times$ volume

Mass of water = 1.0 $\times$ 355 ml

Mass of water = 355 grams.

Heat emitted by water during melting of ice (Q) = mc$\Delta$T

c = specific heat of water = 4.186 J/g $\rm ^\circ C$

m = mass of water = 355 grams

$\Delta$T = change in temperature = final temperature - initial temperature  

$\Delta$T = 5  $\rm ^\circ C$ - 24  $\rm ^\circ C$

Q = 355 $\times$ 4.186 $\times$ (5 - 24) J

Q = -28194.1 J

Heat emitted by water = (Heat absorbed by ice)

Total heat absorbed by the ice = Heat absorbed in the fusion of ice at 0 $\rm ^\circ C$ to water at 0 $\rm ^\circ C$ + heat absorbed in changing from water at 0 $\rm ^\circ C$ to water at 5$\rm ^\circ C$.

Let the mass of ice  = m

Heat absorbed in the fusion of ice at 0 $\rm ^\circ C$ to water at 0 $\rm ^\circ C$ = mass $\times$ latent heat of fusion of ice

= m $\times$ 334 J/g

Heat absorbed in changing from water at 0 $\rm ^\circ C$ to water at 5$\rm ^\circ C$. = mc$\Delta$T

= m $\times$ 4.1 $\times$ 5

= m $\times$ 20.5 J/g

Heat emitted by water = Heat absorbed by ice

-28194.1 J = 334m + 20.5 m

m = 79.4 grams.

Thus the mass of ice that results to lower the temperature of water from 25$\rm ^\circ C$ to 5$\rm ^\circ C$ has been 79.4 grams.

For more information about the change in temperature, refer to the link:

https://brainly.com/question/13945683