Question

The following physical constants are for water, H2O.
The specific heat capacity of the solid = 2.09 J/g oC
The specific heat capacity of the liquid = 4.18 J/g oC
The specific heat capacity of the vapor = 2.09 J/g oC
∆Hfus = 6.02 kJ/mol; ∆Hvap = 40.7 kJ/mol Freezing point = 0.0oC; Boiling point = 100.0oC
How much heat(in kJ) is required to warm 10.0 grams of ice at -5.0oC to a temperature of 70.0oC?

Answer 1

Answer:

$Q\approx6.4~kJ$

Explanation:

Quantity of heat required by 10 gram of ice initially warm it from -5°C to 0°C:

$Q_1=m.C_s.\Delta T$

here;

mass, m = 10 g

specific heat capacity of ice, $C_s=2.09~J.g^{-1}.^{\circ}C^{-1}$

change in temperature, $\Delta T=(5-0)=5^{o}C$

$Q_1=10\times2.09\times 5$

$Q_1=104.5~J$

Amount of heat required to melt the ice at 0°C:

$Q_2=m.\Delta H_{fus}$

where, $\Delta H_{fus}=6020~J/mol$

we know that no. of moles is = (wt. in gram) $\div$ (molecular mass)

$Q_2=\frac{10}{18} \times 6020$

$Q_2=3344.44~J$

Now, the heat required to bring the water to 70°C from 0°C:

$Q_3=m.C_L.\Delta T$

specific heat of water, $C_L=4.18~J/g/^oC$

change in temperature, $\Delta T=(70-0)=70^oC$

$Q_3=10\times 4.18\times 70$

$Q_3=2926~J$

Therefore the total heat required to warm 10.0 grams of ice at -5.0°C to a temperature of 70.0°C:

$Q=Q_1+Q_2+Q_3$

$Q=104.5+3344.44+2926$

$Q=6374.94~J$

$Q\approx6.4~kJ$