Calculate the final temperature (once the ice has melted) of a mixture made up i

Question

Calculate the final temperature (once the ice has melted) of a mixture made up initially of 75.0 mL liquid water at 29.0 degree C and 7.0 g ice at 0.0 degree C. How many grams of steam at 100 degree C would be required to raise the temperature of 35.8 g solid benzene (C_6H_6) from 5.5 degree C to 45.0 degree C? Assume that heat is only transferred from the steam (and not liquid water) and that the steam/water and benzene are separated by a glass wall and do not mix. (The melting point of benzene is 5.5 degree C; Delta H_fus for benzene is 9.87 kJ/mol; specific heat for benzene is 1.63 J/g middot degree C; Delta H_vap for steam at 100 degree C is 40.7 kJ/mol.)

Explanation / Answer

3) mass of water = volume x density

mass of water = 75 x 1 = 75 g

now

Lets assume all the water is brought to 0 C

then

heat released (Q) = m x s x dT

Q = 75 x 4.184 x ( 29-0)

Q = 9100.2 J

now

this energy will be used to melt the ice

heat required to melt the ice = mass of ice x latent heat of fusion

heat required to melt the ice = 7 x 333.55

heat required to melt the ice = 2333.45 J

now

Heat remaining = 9100.2 - 2333.45

heat remaining = 6766.75 J

now

this heat will be used by both the water and melted ice to increase the temperature

after melting 7 g of ice will turn to 7 g of water

so

new total mass of water = 75 + 7 = 82 g

now

6766.75 = 82 x 4.184 x ( T - 0)

T = 19.72

so

the final temperature of the mixture is 19.72 C

4)

moles of C6H6 = 35.8 g / 72 g mol-1

moles of C6H6 = 0.459 mol

now

Q1= heat required to melt solid C6H6 to liquid

Q1 = 0.459 mol x 9.87 kJ/mol

Q1 = 4.53033 kJ

now

Q2 = heat required to bring C6H6 liquid from 5.5 C to 45 C

Q2 = 35.8 x 1.63 x ( 45 - 5.5)

Q2 = 2304.98 J

Q2 = 2.30498 kJ

now

total heat required = Q1 + Q2

Q = 4.53033 + 2.30498

Q = 6.83531 kJ

now

moles of steam required = heat required / dHvap

moles of steam required = 6.83 kJ / 40.7 kJ mol-1

moles of steam required = 0.1679 mol

now

mass of steam required = 0.1679 mol x 18 g /mol

mass of steam required = 3 g

so

3 grams of steam is required

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