Answer:
[tex]m_{iron}=32.1g[/tex]
Explanation:
Hello!
In this case, since the interaction between hot iron and cold water allows the heat transfer from iron to water and therefore we can write up the following energetic equation:
[tex]Q_{iron}+Q_{water}=0[/tex]
Whereas the heat terms can be written in terms of mass, specific heat and temperature change:
[tex]m_{iron}C_{iron}(T_f-T_{iron}) + m_{water}C_{water}(T_f-T_{water}) = 0[/tex]
So we solve for the mass of iron as follows:
[tex]m_{iron} = \frac{m_{water}C_{water}(T_f-T_{water})}{C_{iron}(T_f-T_{iron}) }[/tex]
Now, we plug in the given data to obtain:
[tex]m_{iron} = \frac{-100g*0.449\frac{J}{g\°C} (25.7\°C-22.0\°C)}{0.449\frac{J}{g\°C} (25.7\°C-93.3\°C) }[/tex]
[tex]m_{iron}=32.1g[/tex]
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