Question

I have a question on heat capacity .like i dont know which state. it first has that Specific HEAT CAPACITY of water =4200
heat capacity of ice is 2100
fusion of ice=334 000
then it asked
1.)2.58 kg of water is transformed into ice at -15deg Calculate the amount of heat energy needed to change the water at 60 deg c into water at 0deg c,

and the amount of heat energy needed to change the water at 0 deg c into ice at 0 deg C.

The amount of heat energy needed to change the ice at 0 deg c into ice at -15 deg C

Answer 1

I think you have missed out a bit of the question

Does it say 2.5kg of water at 60degC is transformed into ice at -15degC ?

Answer 2

yes

Answer 3

You should also include the UNITS with the figures you give - they will help you use the correct formula

Answer 4

There are 3 parts to this process:

Cooling water from 60 deg to 0 deg
Freezing water from liquid to ice at 0deg
Coolin ice from 0 deg to -15 deg

You have the necessary constants and mass and change in temp to calculate these 3 changes

Answer 5

the 1st 1 i would have to use specific heat fusion of ice ,second capacity heat capacity,3rd shcapacity of water .

Answer 6

i got 334 000 x 2.58 kg for the second one

Answer 7

Specific heat capacity is defined as the amount of heat required to raise the temperature of one gram of a substance one degrees Celsius (in metric units).

Thermal energy is expressed just like any other energy, with units of Joules.

Therefore, the specific heat of a substance has the following units. \[{J \cdot g \over \Delta T}\]

The values you've been given for specific heat capacity are \[{J \cdot kg \over \Delta T}\]

The amount of energy we need to input into the substance is expressed in the following equation, \[H = m \cdot c \cdot \Delta T\] where m is the mass of the substance, c is the specific heat, and DeltaT is the change in the temperature of the substance.

When a phase change occurs, we have no change in temperature. We are simply adding enough energy to the substance to break the intermolecular bonds (in the case of melting ice).

The heat we need to input into the substance to change the state is given as\[H = m \cdot h_f\] where m is the mass and \(h_f\) is the heat of fusion.

Heat of fusion has units of\[J \cdot kg\]

Three things are happening in this problem.

First, we need to remove enough energy (or heat) from the water to drop its temperature from 60C to its freezing point of 0C.

Then we need to remove enough energy (or heat) from the water to freeze the water. There is no temperature change here.

Finally, we need to cool the ice from 0C to -15C.

We can simply add these three values together to find the total heat we are required to remove.

\[H_T = m \cdot c_w \cdot (0 - 60) + m \cdot h_f + m \cdot c_i \cdot (-15 - 0)\]

\(c_w\) is the specific heat of water and \(c_i\) is the specific heat of ice.

Answer 8

ok this is what i got.
1.)Heat capacity of h20=4200x60x100=25200000
2.)334000x2.58kg=861720
3.)heat capacity of ice=2100x258=5418

Answer 9

Where do you get 100 from in the first equation?

Be careful with your terminology. You're not calculating the heat capacity of the water, you're calculating the heat removed from the water to drop the temperature. Heat capacity is an extensive property of matter.

Answer 10

i have an example in my notesexample in my notes notes

Answer 11

@csmith035
You need to read the help that has been given above, and apply it to your particular question.
Eashmore really has given you the answer - but you didn't apply it

The heat removed when there is no change of state = mass x c x delta t
SO for the first and third parts you need to put the correct values in for m, c and delta t

Heat removed when there IS a change of state = mass x hf

SO for the second part you need to put the correct values for mass and hf

THEN add them all together

Note that the correct value is negative because you are REMOVING heat. (T2-T1 <0)