Question

calculate the amount of heat (in kJ) required to warm 5.00 g of ice at -10 C to steam at 110 C

Answer 1

Step 1: The energy required to raise the temperature of ice from -10 ^oC to its melting point i.e. 0 ^oC.
   Q₁ = m_ice x C_p,ice x T
= (5 g) x (2.108 J/ g-^oC) x (10 ^oC)
= 105.4 J
Step 2: The energy required to convert ice to liquid water at 0 ^oC.
   Q₂ = m_water x H_fus
= (5 g) x (333.55 J/ g)
= 1667.75 J
Step 3: The energy required to raise the temperature of H₂O(l) from 0 ^oC to its boiling point at 1 atm i.e. 100 ^oC.
   Q₃ = m_water x C_p,water x T
= (5 g) x (4.184 J/ g-^oC) x (100 ^oC)
= 2092 J
Step 4: The energy required to convert H₂O liquid to steam at 100 ^oC.
   Q₄ = m_water x Hvaporization
= (5 g) x (2230 J/ g)
= 11150 J
Step 5: The energy required to raise the temperature of steam from 100 ^oC to 110 ^oC.
   Q₅ = m_water x C_p,steam x T
= (5 g) x (1.996 J/ g-^oC) x (10 ^oC)
= 99.8 J
Therefore total energy required will be Q = Q₁ + Q₂ + Q₃ + Q₄ +Q₅.
= 105.4 + 1667.75 + 2092 + 11150 + 99.8 J
   =15114.95 J
   = 15.115 kJ
Kindly note that the values of C_p,ice , C_p,water , C_p,steam , H_fusion and H_vaporization are taken from the internet and to completely match the solution kindly use the values given in the table provided to you.