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A blacksmith drops a 550 °F piece of iron into a vat of 75 °F water in order to cool it to 100 °F. How many kilograms of water are needed per kilogram of iron? Assume all the thermal energy from the iron is transferred to the water and none of the water evaporates. The specific heats of water and iron are 4186 J/kg×°C and 448 J/kg×°C, respectively.

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Answer:

1.93 kg water/kg iron

Explanation:

All the thermal energy from the iron is transferred to the water. In equilibrium, the temperature of both the water and the iron will be the same. The heat that an object loses or gains after a change in value in its temperature is equal to:

[tex]Q = mc*(T_f-T_o)[/tex]

Then,

[tex]-Q_{iron} = Q_{water}[/tex]

Before solving the problem, let's convert the values of temperature to Celsius:

(550°F -32)*5/9 = 287.78 °C

(75°F - 32)*5/9 = 23.89 °C

(100°F -32)*5/9 = 37.78°C

Now, we can solve:

[tex]-Q_{iron} = Q_{water}\\m_{iron}*c_{iron}*(T_o_i-T_f_i) = m_{water}*c_{water}*(T_f_w-T_o_w)\\\frac{m_{water}}{m_{iron}} = \frac{c_{iron}(T_o_i-T_f_i)}{c_{water}(T_f_w-T_o_w)} =\frac{448J/kg^oC(287.78^oC - 37.78^oC)}{4186J/kg^oC(37.78^oC-23.89^oC)}= 1.93 kg_{water}/kg_{iron}[/tex]

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