Answer
The final temperature is $123.7^{\circ}C$
Work Step by Step
Use the equation $q=mcΔT$, where q represents the heat that flows into a system to increase its temperature, m represents the mass of the substance, 'c' represents the heat capacity of the substance and ΔT represents the change in temperature. We can rearrange and solve for the change in temperature, (i.e. ΔT)
First we must convert the heat unit from cal to J by multiplying by 4.184 J because $1 cal = 4.184 J$ so $146 cal \times 4.184 J = 610.8 J$. Also the heat capacity of lead is 0.128 as found online.
$q=mcΔT$
$ΔT=\frac{q}{mc}$
$ΔT=\frac{610.8 J}{57 g\times 0.128 }$
Therefore, the temperature change is $83.7^{\circ}C$ and since $\Delta T = T_{final} - T_{initial}$, we can solve for the final temperature since we know the initial temperature to be $47^{\circ} C$ as its given in the question. Thus, $T_{final} = \Delta T +T_{initial}$ so $83.7^{\circ}C + 47^{\circ}C = 123.7^{\circ}C$
