Question

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.

Answer 1

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,

Q_ice = - Q_water

Q_ice = Q2 + Q3

Q_water = Q1

Therefore,

Q1 = Q2 + Q3

29727.7 = 334m + 25.122m

m = 82.78 g