How to calculate the mass of iron with a temperature of 325 °C needed to melt ice and raise the temperature of water from 0 to 22 °C?


> 
> A large well-insulated container holds a mixture of $75\ \mathrm g$ of ice and $100\ \mathrm g$ of water at $0\ \mathrm{^\circ C}$. Using the data given below, calculate the mass in grams of solid iron, $\ce{Fe}$, at $325\ \mathrm{^\circ C}$ that you would have to add to this mixture in order to melt *all* of the ice and raise the temperature of the resulting $175\ \mathrm g$ of water to $22\ \mathrm{^\circ C}$.
> 
> 
> Specific heat capacity of *water* is $c\_p(\ce{H2O})=4.184\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$  
> 
> Specific heat capacity of *iron,* $\ce{Fe}$, is $c\_p(\ce{Fe})=0.45\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$  
> 
> Molar enthalpy of fusion (melting) of water is $\Delta\_\text{fus}H\_\mathrm m(\ce{H2O})=6.01\ \mathrm{kJ\ mol^{-1}}$  
> 
> $M(\ce{H2O})=18\ \mathrm{g\ mol^{-1}}$
> 
> 
> 


My answer was that you needed $25041.67\ \mathrm J$ to melt the ice and $16108.4\ \mathrm J$ to raise the temperature to 22 degrees Celsius.


Therefore $mc \Delta T(\ce{Fe}) = -(25041.67\ \mathrm J + 16108.4\ \mathrm J)$ and dividing this by the mass of iron and the change in temperature results in a mass of $301.8\ \mathrm g$.


Is this correct?




I can not see anything wrong with your calculations at all. For posterity, the following is a rough method to confirm:


**Melting ice**


Enthalpy of fusion $\Delta H\_\text{ice} = 6.01\ \mathrm{kJ\ mol^{-1}}$  

Mass of ice $m = 75\ \mathrm g$  

Molar mass of $\ce{H2O}$ $M = 18\ \mathrm{g\ mol^{-1}}$ 


*Hint: usually, when you are given mass and molar mass, then the amount of the substance is required.*


Note that the Enthalpy of fusion ($\Delta H\_\text{ice}$) is measurement in $\mathrm{kJ\ mol^{-1}}$, and as the ice is measured in grams, and you are provided with the molar mass, you'll need to use the formula:


$$n = \frac{m}{M}$$


hence:


$$n\_\text{ice} = \frac{75\ \mathrm g}{18\ \mathrm{kJ\ mol^{-1}}}$$


$$n\_\text{ice} = 4.1666\ \mathrm{mol}$$


Therefore, the enthalpy of fusion:


$$\Delta H\_\text{fus} = 6.01\ \mathrm{kJ\ mol^{-1}} \times 4.1666\ \mathrm{mol} = 25.04167\ \mathrm{kJ} = 25041.67\ \mathrm J$$


**Heating water**


Specific heat capacity of water $c\_p = 4.184\ \mathrm{J\ g^{-1}\ ^\circ C}$  

Mass of water $m = 175\ \mathrm g$  

Temperature change $\Delta T = 22\ \mathrm{^\circ C}$ 


So:


$$Q = mc\_p \Delta T$$


$$Q = 175\ \mathrm g \times 4.184\ \mathrm{J\ g^{-1}\ ^\circ C} \times 22\ \mathrm{^\circ C} = 16108.4\ \mathrm J$$


**Total energy used by the water/ice mixture**


$$Q\_\text{total} = 25041.67\ \mathrm J + 16108.40\ \mathrm J = 41150.07\ \mathrm J$$


**Iron**


Specific heat capacity of iron $c\_p = 0.45\ \mathrm{J\ g^{-1}\ ^\circ C}$  

Mass of water $m = X$ (*unknown*)  

Temperature change $\Delta T = 325\ \mathrm{^\circ C} - 22\ \mathrm{^\circ C} = 303\ \mathrm{^\circ C}$ decrease 


As, according to conservation laws:


$$Q\_{\ce{H2O}} = Q\_{\ce{Fe}}$$


Then:


$$41150.07\ \mathrm J = X \times 0.45\ \mathrm{J\ g^{-1}\ ^\circ C} \times 303\ \mathrm{^\circ C}$$


Or:


$$41150.07\ \mathrm J = 136.35\ \mathrm{J\ g^{-1}}\times X$$


Then:


$$X = \frac{41150.07\ \mathrm J}{136.35\ \mathrm{J\ g^{-1}}}$$


$$X = 301.8\ \mathrm g$$


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