1
$\begingroup$

enter image description here

My new try: Given data

Energy given by reaction =1690 kJ

Initial temperature = 0 oC

Final temperature = 50 oC

C water = 4.18 Jmol-1 K-1

?Hfus ice = 6.01 kJ*mol-1

Calculate Mass of ice = ?

Ice melts at constant temperature of 0 degree C and changes the phase from solid to liquid .

Then liquid water absorbs the energy and changes from 0 degree C to 50 degree C

Change in temperature of the water is ?T = (50 oC - 0 oC )= 50 oC

Let’s set up the equation as follows to calculate the mass of ice

q= (m*?Hfus) + (m*Cwater * ?T)

where , q = energy , m = mass , ?T = (Tf-Ti)

convert given energy from unit kJ to Joules

(1690 kJ * 1000 J)/1kJ = 1690000 J

Convert ?H fus from kJmol-1 to Jg-1

6.01kJ*mol-1 (1000 J/1kJ) = 6010 Jmol-1

6010 Jmol-1 / 18.0148 gmol- = 333.6 J *g-1

Convert C water from J per mol to J per g

4.18 Jmol-1 oC /18.0148 g = 0.232 J*g-1 oC-1

Now lets put the values in the formula and calculate the mass of ice

q= (m*?Hfus) + (m*Cwater * ?T)

1690000 J = (m X 333.6 Jg-1) + (m X 0.232 Jg-1 *oC-1 * 50 oC )

1690000 J = m X 345.2 J*g-1

1690000 J /345.2 J*g-1 = m

4895.7 g = m

Please help me!!

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Most of your manipulations seem correct, but I think the question has inaccurate data. The specific heat of water is $4.18 \ \mathrm{J\ \mathbf{g^{-1}}\ K^{-1}}$, not $4.18 \ \mathrm{J\ mol^{-1}\ K^{-1}}$. $\endgroup$
    – Nicolau Saker Neto
    Jun 9, 2015 at 12:32
  • $\begingroup$ My teacher said that the information is correct $\endgroup$
    – Josh
    Jun 9, 2015 at 12:45
  • 3
    $\begingroup$ @NicolauSakerNeto is right, 1 mole of water contains ~20 grams, so using the wrong units will give you an answer that's off from the real answer by a factor of ~20. (However maybe it's wrong in the Homework's answer as well, which seems to happen sometimes) $\endgroup$
    – Dan
    Jun 9, 2015 at 14:35

1 Answer 1

Reset to default
2
$\begingroup$

Energy needed for melting and heating up the water

$$ Q = \Delta H_{\text{fus}} \cdot m + m \cdot c_p \cdot \Delta T $$

solve for $m$:

$$ m = \frac{Q}{\Delta H_{\text{fus}} + c_p \cdot \Delta T } $$

The molar heat capacity for water is around $75\,\frac{\text{J}}{\text{mol K}}$. You've been given the specific heat capacity which is around $4.18\,\frac{\text{kJ}}{\text{kg K}}$. All that's left to do is convert the enthalpy of fusion to $\frac{\text{kJ}}{\text{kg}}$ and you can solve the equation.

$$ \Delta H_{\text{fus}} = \frac{6.01\,\frac{\text{kJ}}{\text{mol}}}{18.02\,\frac{\text{g}}{\text{mol}}} = 0.3335\,\frac{\text{kJ}}{\text{g}} = 333.5\,\frac{\text{kJ}}{\text{kg}} $$

Calculate $m$:

$$ m = \frac{1690\,\text{kJ}}{333.5\,\frac{\text{kJ}}{\text{kg}} + 4.18\,\frac{\text{kJ}}{\text{kg K}} \cdot 50\,\text{K}} = 3.12\,\text{kg} $$

The combustion of 1 mol of propane supplies enough heat to melt 3.12$\,$kg of ice and and heat the resulting water up to 50$\,$°C

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.