Question

Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of H2O is 273 K. Other useful information about water is listed below. Cp,liquid = 4.184 J/(g•K) Cp,solid = 2.093 J/(g•K) ΔHfusion = 40.7 kJ/mol

Answer 1

Answer : The amount of heat released, 45.89 KJ

Solution :

Process involved in the calculation of heat released :

$(1):H_2O(l)(314K)\rightarrow H_2O(l)(273K)\\\\(2):H_2O(l)(273K)\rightarrow H_2O(s)(273K)\\\\(3):H_2O(s)(273K)\rightarrow H_2O(s)(263K)$

Now we have to calculate the amount of heat released.

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

Q = amount of heat released = ?

m = mass of water = 27 g

$c_{p,l}$ = specific heat of liquid water = 4.184 J/gk

$c_{p,s}$ = specific heat of solid water = 2.093 J/gk

$\Delta H_{fusion}$ = enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole

conversion : $0^oC=273k$

Now put all the given values in the above expression, we get

$Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]$

$Q=45896.798J=45.89KJ$     (1 KJ = 1000 J)

Therefore, the amount of heat released, 45.89 KJ

Answer 2

1.50 mol and -66.2kJ