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?

Question

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?

Answer 1

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:

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

$$X = 301.8\ \mathrm g$$