Question

the average molar heat capacity of ice and water are 37.8 and 75.6 J/mol. enthalpy of fusion of ice is 6.012 KJ/mol. amount of heat required to change 10g of ice at -10 degree Celsius to water at 10 degree celsius would be
(1) 2376J
(2) 4752 J
(3) 3970 J
(4) 1128 J

Answer 1

Hello there,

● Answer-
H = 3970 J

● Explaination-
n1 = 10 g = 10/18 = 0.555 mol
n2 = 10 g = 10/18 = 0.555 mol
∆Hf = 6020 J/mol
c1 = 37.6 /molK
c2 = 75.2 /molK

# Solution-
To convert ice into water three enthalpies are involved.

Heat required to change temp of ice from -10 to 0 -
∆H1 = n1c1∆T
∆H1 = 0.555 × 37.6 × 10
∆H1 = 208 J

Heat required to change temp of water from 0 to 10 -
∆H2 = n2c2∆T
∆H2 = 0.555 × 75.2 × 10
∆H2 = 418 J

Enthalpy of melting-
∆Hf = 0.555 × 6020
∆Hf = 3344 J

Total heat required is-
H = ∆H1 + ∆Hf + ∆H2
H = 208 + 3344 + 418
H = 3970 J

Therefore, heat required to convert given ice to water is 3970 J.

Hope it helps...

Answer 2

"Given,

The molar heat capacity of Ice, $c_1= 37.8kJ/mol$

The molar heat capacity of water, $c_2 = 75.6J/mol$

The enthalpy fusion of ice = 6.012KJ/mol

Consider the following equation,

$n_1 = 10 g = 10/18 = 0.555 mol$

$n_2 = 10 g = 10/18 = 0.555 mol$

∆Hf = 6020 J/mol

$c_1 = 37.6 /molK$

$c_2 = 75.2 /molK$

i) Heat required to change temperature of ice from -10degree to 0degree

$\Delta H_{ 1 }=\quad n_{ 1 }C_{ 1 }p\Delta T\quad =\quad 0.55\quad \times \quad 37.8\quad \times \quad (0-(-10))\quad =\quad 208J$

ii) Enthalpy of fusion

$\Delta H_{ 2 }=\quad \Delta H_{ fusion\quad}=\quad 6.020KJ\quad =\quad 6020J\quad =\quad 6020\quad \times \quad 0.55=\quad 3311J$

iii) Heat required to change temperature from 0 degree to 10 degree

$\Delta H_3= n_2C_2p\Delta T = 0.55 x 75.6 x(10-0) = 415J$

Total heat required $H = \Delta H_1 + \Delta H_2 + \Delta H_3$

                                                H= 3726J "