How much heat is required to bring 500g of ice from -20 C to completely boiled away at 100 C?

Question

And please explain it too. Thanks!

Answer 1

we calculate it in 4 steps. Iuse as a thermal unit the calory which is 4.18J

First you must bring the ice from-20 to 0°c for this the formula is
Q = mc(tf-ti) m =500 c 0.5 (tf-ti) =20

you need Q1= 500*20*0.5 = 10000cal

then you must melt ice Q2 =m *Hm hm = 80 cal/g

so Q2 = 80*500 =40000cal

thirs bring the water from 0 to 100°

Q3 = m c (100-0) = 500*1*100= 50000cal

and Fourth Step evaporate the water

Q4 = m He= 500*540 = 270000cal

the total amount is
Q1+Q2+Q3+Q4= 10000+40000+50000+270000= 370000cal
in joules 370000*4.185=1548500J

Answer 2

There are 4 parts to your problem

In part 1 you have to warm the ice from -20 to its melting point at zero.

The formula is q = m Cp delta T

Please note that the Cp of ice is not the same as for water. You need to look it up. Its about 2 J/g oC

In part 2. you have to melt the ice.

The formula is q = m Hf. Hf is the heat of fusion of ice. Its about 338 J / g . Look up the exact value. (I think it is exactly 6.02 J / mol)

In part 3 you will heat the water from 0 to 100 0C.

Use q= m Cp delta T. Now the Cp is 4.184 J /g oC

In part 4 you will boil the water.

The formula is q = m Hv Heat of vaporization. You need to look up this value.

Add the q of all 4 parts together and you are done

Answer 3

use this equation Q=(m)(c)(delta t)
Q is mols times heat of fusion (Hf) (which is 6008 jouls)
m is mass (grams)
c is change in temp. which is 4.184 Jouls
t is temperature change, from -20 to 0 degrees, so t is -20
then use q=(mol)(Hf)
then q-mcdeltat again only for t put 100 degrees
add all of the formula answers together and that will give you temp. in Joules