Final temperature when 75 g ice melts in 330 g water at 22°C

[jinx007]

PLEASE HELP ME TO ANSWER THIS QUESTION I AM STUCK

An ice cube of 75 g is drop into a glass containing 330 g of water at a temperature of 22 ' Celsius. The ice is at 0 ' Celsius and melts so that the temperature of the water decreases.

Calculate the final temperature of the water when all the ice has melted. Assume that no heat lost to the glass or the surrounding.

Specific latent heat of fusion of ice = 3.3 x 10^5 J/kg

specific heat capacity of water = -4.2 x 10^3 J/kg K

Please help me i don't really know help to proceed

 

[Doc Al]

Call the final temperature X.

How much energy is required to melt the ice?

How much energy is required to heat the melted ice water from 0 degrees to X?

How much energy must be removed from the warm water to bring it down from 22 degrees to the final temperature?

 

[jinx007]

Call the final temperature X.

How much energy is required to melt the ice?

How much energy is required to heat the melted ice water from 0 degrees to X?

How much energy must be removed from the warm water to bring it down from 22 degrees to the final temperature?

Energy to melt ice to water at 0 ' Celsius = q = m lf



Energy to melt ice to water from 0 to X using q = mcQ
0.075 x -4.2 x 10^3 x( X - 0) The last part i cannot understand..please check whether i am on the right track, please explain to me a bit more in detail

 

[Doc Al]

Yes, you're on the right track. Hint: The energy you must add to the ice to melt it and then heat it up must equal the energy removed from the warm water.

You'll set up an energy equation and then solve for X.

 

[jinx007]

Yes, you're on the right track. Hint: The energy you must add to the ice to melt it and then heat it up must equal the energy removed from the warm water.

You'll set up an energy equation and then solve for X.

"""...How much energy must be removed from the warm water to bring it down from 22 degrees to the final temperature?..."""

How should i calculate the energy that must be removed from the warm water to bring it from 22 to X

should i use q = mcQ
0.33 x -4.2 x 10^3 x ( 22 - Q )

by the way the upper part i understand.

 

[Doc Al]

How should i calculate the energy that must be removed from the warm water to bring it from 22 to X

should i use q = mcQ
0.33 x -4.2 x 10^3 x ( 22 - Q )

The equation you want is the same one that you used to find the heat you need to add to warm the ice water:

Q = mcΔT = mc(22 - X)

 

[jinx007]

The equation you want is the same one that you used to find the heat you need to add to warm the ice water:

Q = mcΔT = mc(22 - X)

So if i round up all what i understand the answer should be like that:

m lf + 0.075 x -4.2x10^3 x (x-0) = 0.33 x -4.2 x 10^3 (22-x)

when i solve i will have the value of X that is the temperature

 

[Doc Al]

So if i round up all what i understand the answer should be like that:

m lf + 0.075 x -4.2x10^3 x (x-0) = 0.33 x -4.2 x 10^3 (22-x)

when i solve i will have the value of X that is the temperature

Looks good. (But get rid of the minus sign in front of the specific heat.) Don't forget to plug in the values for m and lf.