When the heat is absorbed by the ice then ice will melt to form liquid water

![enter image description here](https://i.stack.imgur.com/4IAvr.jpg)


My new try:
Given data


Energy given by reaction =1690 kJ


Initial temperature = 0 oC


Final temperature = 50 oC


C water = 4.18 J*mol-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 kJ*mol-1 to J*g-1


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


6010 J*mol-1 / 18.0148 g*mol- = 333.6 J \*g-1


Convert C water from J per mol to J per g


4.18 J*mol-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 J*g-1) + (m X 0.232 J*g-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!!




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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