Question

The enthalpy change for converting 10.0 g of ice at -25.0°C to water at 80.0°C is __________ kJ. The specific heats of ice, water, and steam are 2.09 J/g·K , 4.184 J/g·K ,  and 1.84 J/g·K  respectively. For H₂O, ΔH_fus = 6.01 kJ/mol, and ΔH_vap =40.67 kJ/mol

Answer 1

Answer –

We are given, mass of ice = 10.0 g , ti = -25^o C, tf = 80.0 ^oC,

The specific heat of idea, Ci = 2.09 J/g·K , specific heat for water Cw = 4.184 J/g·K

ΔH_fus = 6.01 kJ/mol =6.01*10³ kJ/mol

ΔH_vap =40.67 kJ/mol = 4.067*10⁴ kJ/mol

Step 1) Calculate the heat from -25^o C to 0.00^oC

We know formula

q1 = m*Ci*Δt

   = 10.0 g * 2.09 J/g·K * (0.0-(-25))^oC

   = 522.5 J

Step 2) Calculate the heat from 0.00 ^o C to 0.00^oC

We need to first calculate moles of 10.0 g water, because ΔH_fus in kJ/mol

Moles of water = 10.0 g / 18.015 g.mol^-1 = 0.555 moles

q2 = m* ΔH_fus

     = 0.555 mol *6.01*10³ J/mol

    = 3336.1 J

Step 3) Calculate the heat from 0.00 ^o C to 80.00^oC

q3 = m*Cw*Δt

     = 10.0 g * 4.184 J/g·K * (80.0-0.00)^oC

     = 3347.2 J

Step 4) Total heat from -25^o C to 80.0 ^oC

q = q1 +q2 +q3

   = 522.5 J + 3336.1 J + 3347.2 J

= 7206 J

   = 7.206 kJ

So change in enthalpy is the heat required for the melting ice to water and it is 7.206 kJ