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.
Please show all work

Answer 1

Let the mass of the ice be m = 1.00 kg

We have to solve the problem stepwise

The data required for it is as follows

Specific heat of ice is S_ice = 2.1 * 10³ J/kg K

Latent heat of fusion of ice is L_ice = 3.35 * 10⁵ J/kg

Specific heat of water is S_w = 4.2 * 10³ J/kg K

Latent heat of vaporization L_v = 2.26 * 10⁶ J/kg

Specific heat of steam S_steam= 2.08 * 10³ J/kg K

1)Heat required to raise the temperature of from -10⁰ C to 0⁰ C

             Q1 = m * S_ice * ∆T = 1 * 2.1 * 10³ * ( 0 – ( -10 ) )

                   = 21000 J

2)Heat required to convert ice at 0⁰ C to water at 0⁰ C

                Q2 = m * L_ice = 1 * 3.35 * 10⁵ = 335000 J

3) Heat required to raise the temperature of water to 100⁰ C

                  Q3 = m * S_w * ∆T = 1 * 4.2 * 10³ * ( 100 – 0 )

                        = 420000 J

4) Heat required to convert water at 100⁰ C to steam at 100⁰ C

                  Q4 = m * L_v = 1 * 2.26 * 10⁶ = 2260000 J

5) Heat required to raise the temperature of stem from 100⁰ C to 110⁰ C

                 Q5 = m * S_steam * ∆T = 1 * 2.08 * 10³ * ( 110 – 100 )

                       = 20800 J

Now total heat supplied is Q = Q1 + Q2 + Q3 + Q4 + Q5

                                                      = 21000 + 335000 + 420000 + 2260000 + 20800

                                                     = 3056800 J

b) 1 Joule = 0.23 cal

so 3056800 J = 3056800 * 0.23 cal = 703064 cal = 703.064 kcal

So 703.064 kcal is supplied to convert ice to steam