Question

In: Chemistry

How much ice (in grams) would have to melt to lower the temperature of 355 mL...

How much ice (in grams) would have to melt to lower the temperature of 355 mL of water from 26 ∘C to 6 ∘C? (You must consider the enthalpy in cooling the water, as well as the enthalpy in melting the ice and the enthalpy to heat the melted ice from 0∘C up to the system's final temperature of 6 ∘C. Also assume that the density of water is 1.0 g/mL.) Express your answer using two significant figures.

Solutions

Expert Solution

Given -

The volume of water = 355 mL

Since , the density of water is 1g/ mL

Density =mass / volume

mass = density * volume

mass = 1g/mL * 355 mL

mass = 355g

The enthalpy of cooling 350 ml of water from 260 C to 60 C, is given by Q1-

Q1 = m * C * ΔT

Q1 = heat released

m = mass of water

C = specific heat of water ( 4.187 J/g 0 C )

ΔT = change in temperature ( final temperature – initial temperature )

Q1 = 355g * 4.187 J/g 0 C ( 60 C - 260 C )

Q1 = - 29727.7 J

This energy (Q1) will be equal to the energy required to melt the ice and raise the temperature of the ice from 0°C to 6°C.

Let the mass of ice required be 'm'.

The enthalpy change (enthalpy of fusion) for the conversion of ice at 00 C to water at 00 C , is given by Q2-

Q2 = m*delta H (fusion)

m= mass of ice

delta H (fusion) = enthalpy change for the conversion of ice to water ( 334 J/g )

Q2 = m * (334 J/g)

Q2 = 334 m

The enthalpy change for the conversion of water at 0 0 C to 6 0 C , is given by Q3 -

Q3 = m * C * delta T

Q3 = m * 4.187 J/g 0 C ( 60 C - 00 C )

Q3 = 25.122m

In the system,

Qice = - Qwater

Qice = Q2 + Q3

Qwater = Q1

Therefore,

Q1 = Q2 + Q3

29727.7 = 334m + 25.122m

m = 82.78 g


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