= Solution
<Specific heat capacity> is <heat capacity> per unit <mass>. For a specified thermodynamic process without a <phase transition>,
$$
\boxed{c=\frac1m\frac{dQ}{dT}.}
$$
The constraint matters: constant-<pressure> and constant-volume <heat capacities> need not coincide. Here the supplied iron value and water value are treated as constants.
At the stipulated barometer reading, iron left in boiling water reaches $100^\circ\mathrm C$. Taking water's <specific heat capacity> as one in the given units, the iron has <heat capacity> $100(0.114)=11.4$. The adhering gramme of hot water adds another one. Thus the hot material has total <heat capacity> $12.4$, compared with $150$ for the initially cool water.
The no-loss <calorimetry> <energy> balance is
$$
100(0.114)(100-T)+1(100-T)=150(T-10),
$$
so
$$
\boxed{T=\frac{12.4(100)+150(10)}{162.4}
=\frac{3425}{203}\,{}^\circ\mathrm C\simeq16.8719^\circ\mathrm C.}
$$
The adhering water enters on the hot side, not as extra water initially at $10^\circ\mathrm C$. The vessel's <heat capacity> is neglected, as none is supplied.
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