Question

Heat is added to 1.07 kg of ice at -20 °C. How many kilocalories are required to change the ice into steam at 140 °C ??

Please show all work, thanks!!

Answer 1

m = 1.07 kg

Step 1: Heat required to raise the temperature of ice from -20 °C to 0 °C Use the formula

q = mcΔT

where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature

q₁ =1.07*10^3*2.09*(0 - -20) = 44726 J

Step 2: Heat required to convert 0 °C ice to 0 °C water

q₂ = m·ΔH_f = 1.07*10^3*334 =357380 J

Step 3: Heat required to raise the temperature of 0 °C water to 100 °C water

q₃ = mcΔT = 1.07*10^3*4.18*(100-0) =447260 J

Step 4: Heat required to convert 100 °C water to 100 °C steam

q₄ = m·ΔH_v = 1.07*10^3*2257 =2414990 J

Step 5: Heat required to convert 100 °C steam to 140 °C steam

q₅ = mcΔT =1.07*10^3*2.09*(140-100) =89452 J

total heat = q1 +q2+q3+q4+q5

= 44726 +357380 +447260 +2414990 +89452 = 3.353808 × 10⁶ J

or 801.5793499  kilocalories