Question

Compute the total quantity of heat (both in Kcal and J) required to raise the temperature of 1.00 kg of ice at -10.0 degrees C to 1.00 kg of steam at 110.0 degrees C. Assume no loss of mass or heat in this process. Note that this will requires several computations that must be completely added up, and you must include all phase changes and use the correct specific heats.

Answer 1

multiply the latent heat of fusion of ice by the mass in kg
which is
0.025* 333.55= 8.33875 KJ= your answer

2 for this we multiply the latent heat of vapourization of the steam by it's mass. step 1

then we multiply the mass of the steam (which is now water) by the temperature change in K or C and by the Specific heat capacity of water. step 2

Then we add this value to the first step 3
which is
(0.11* 2258) + (0.11*20*4.18)
= 248.38 + 9.19600
=257.57600 KJ

looks like you were wrong.

3 for this we first calculate the amount of heat required to change the temperature of the ice to 0
which is
0.1 * 2.108
=0.2108
then calculate how much heat is required to convert the ice to water which is
0.1 *334
=33.4
then calculate ho much heat will convert the water to 100 degrees
=0.1 * 100 * 4.186
=41.86
then calculate how much heat will change the water to steam
which is
0.1*2258
=225.8
After that we calculate how much quantity of heat will change the temperature of the steam to 110
=0.1*10*1.996
=1.996
Now we add them all up
1.996 + 225.8 + 41.86 + 33.4 + 0.2108
which is
=303.2668 KJ = 3.0 x 10^5 = your answer

4 for this one we'll use x as th heat of fusion
so
0.015*x +4.186*9.5*0.015 = 0.1*4.186* (20 - 9.5)
( the equation is because the heat gained by the ice equals the heat lost by the water)
0.015x= 0.1*4.186* (20 - 9.5) - 4.186*9.5*0.015
0.015x = 3.798795
x= 3.798795 / 0.015
x = 253.25300 KJ = 253253J = your answer