All Leaving Cert Physics posts
• #### Specific latent heat of fusion of ice adamskiboy528491

In an experiment to measure the specific latent heat of fusion of ice, a student obtained the following results:

Mass of copper calorimeter = 36g

Mass of calorimeter and water = 80.3g

Initial temperature of water and calorimeter = 19 (degrees)

Temperature of ice = 0 (degrees)

Final temperature of water, calorimeter and ice = 11.5 (degrees)

Specific heat capacity of water = 4180 J (per) kg (per) Kelvin

Specific heat capacity of copper = 390 J (per) kg (per) Kelvin

Find the specific latent heat of fusion of ice (in kilograms)

1. #### Sinott

Firstly, you are missing data. You also need the mass of the ice added.

Also, specific latent heat is measured in kilo*joules*, not kilograms.

Method:

Heat loss = Heat Gain (theoretically, assuming no heat loss to environment)

Heal loss: Calorimeter and Water

Heat Gain: Ice latent heat (melting) and warming now melted ice

Q=(mass)(heat capacity)(change temp) Q=mass(latent heat)

mc(change temp){calorimeter}+mc(change temp){water}=mc(change temp){melted ice} + (mass ice)(heat of fusion) {to find}

Find mass of water by subtraction. Find temp differences by subtraction also

Ok, thank you very much for your time!

(I just wanted to clarify that the grams (in the masses) had to be put into kilograms (for the people who were going to work it out themselves) instead of putting it into grams).

BTW, the was no mention of the mass of ice (in my data). Do you know how to find it, according to this data I got? (All I know is that I'm suppose to subtract it).

4. #### Sinott

I am almost certain that data is missing. Usually in these questions you are given the mass of ice straight away or given final mass of melted ice, calorimeter and original warm water.

You could still solve in terms of m (mass of ice) if you wanted. Do as below but leave mass of ice as m and factorise out, divide across by m at end after subbing in all of the other numbers

here's an example I got from www.thephysicsteacher.ie

Specific heat capacity of water = 4200 J kg–1 K–1

s.h.c. of aluminium = 910 J kg-1 K-1

37. [2002]

In an experiment to measure the specific latent heat of fusion of ice, warm water was placed in an aluminium calorimeter. Crushed dried ice was added to the water.

The following results were obtained.

Mass of calorimeter.......................................= 77.2 g

Mass of water.................................................= 92.5 g

Initial temperature of water...........................= 29.4 0C

Temperature of ice ........................................= 0 C

Mass of ice.....................................................= 19.2 g

Final temperature of water.............................= 13.2 0C

Room temperature was 21 0C.

(i) What was the advantage of having the room temperature approximately halfway between the initial temperature of the water and the final temperature of the water?

(ii) Describe how the mass of the ice was found.

(iii) Calculate a value for the specific latent heat of fusion of ice

(iv) The accepted value for the specific latent heat of fusion of ice is 3.3 × 10^5 J kg-1; suggest two reasons why your answer is not this value.

Solution:

37.

(i) Heat lost to surroundings when the system is above room temperature would cancel out the heat taken in from the surroundings when the system was below room temperature.

(ii) Final mass (of calorimeter + water + ice) - initial mass (of calorimeter + water)

(iii) mcΔθAl + mcΔθwater = mlice +mcΔθmelted ice

Fall in temperature = 16.2 oC

Ans = 3.2 × 105 J kg-1

(iv) Thermometer not sensitive enough, lack of insulation, lack of stirring, heat loss/gain to surroundings, too long for ice to melt, inside of calorimeter tarnished, splashing, heat capacity of thermometer