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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!!

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    $\begingroup$ 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 '15 at 12:32
  • $\begingroup$ My teacher said that the information is correct $\endgroup$ – Josh Jun 9 '15 at 12:45
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    $\begingroup$ @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 '15 at 14:35
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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

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