Question

Find the heat required to converte 1 kg of ice at 0 degree celsius to vapour at 100 dgree celuis specfic hrat capacity of ice =2100kjkg

Answer 1

Answer:

3,027,000 J or 3,027 kJ

Explanation:

I studied this in chemistry although your process may be different.

Formulas used:

q=(mass)(specific heat capacity)(Final temp-Initial temp)
q=(mass)(latent heat)
H=sum up the amounts of q

*q represents heat and H represent enthalpy

First,

In order to find the total amount of heat we need to break it into steps , since water is in different states and temperatures.

We know the freezing and boiling points of water are 0 degrees Celsius and 100 degrees Celsius, respectively. Using this, we can separate the states;

we start with ice (solid) and then water (liquid) and end with steam (gas) so you can think of it as increasing in temperature

*one reason we don't go from solid to gas is because we aren't given latent heat of sublimation

Second,
So we know it goes from ice @ -10 degrees Celsius (or 263.15 K) to steam @ 100 degrees Celsius (or 373.15 K), and we know the freezing and boiling points of water and the formulas; we can now solve the problem!

we need to find the heat (q) needed to raise the temp of ice from -10 degress Celsius to 0 degrees celsius and so we use:

q=(mass)(specific heat capacity)(final temp-initial temp)
mass is 1 kg, specific heat is of ice, final temp is 0 but it is better to convert to Kelvin , and initial temp is -10 degrees Celsius 
you get 21000 J

every time we need to find the heat to RAISE or LOWER the temperature we use this formula

Third, 
Now that we have the ice in 0 degrees Celsius and we have calculated the amount of heat required to bring the ice to this temperature, we can now convert it to liquid using the other formula!

q=(mass)(latent heat of fusion)
you get 3.36 x 10^5 J

Fourth,
the rest is the same process until you get to steam at a temp of 100 degrees Celsius.

Tried to explain it the best I can.
Hope this helps!

Rahul Aryan

Answer 2

’s answer is correct. It is obvious only when you know that the latent heat of fusion of water is 79.7 calories per gram. “Latent heat” is the amount of energy required to change the state of a substance (from liquid to solid, liquid to vapour, in either direction) without changing