Answer
$T_{f} = 149.22 °F $
Work Step by Step
Given:
brass with $m = 10 lb_{m}$
Temperature 1 ($T_{0}$) = 25°C (77°F)
Amount of heat supplied (ΔQ) = 65 Btu
Required:
Final temperature ($T_{f}$)
Solution:
Using the modified form of Equation 19.1:
$ c = \frac{dQ}{dT}$
where $c_{p}$ of brass = $(375 \frac{J}{kg.K}) (\frac{2.39 \times 10^{-4} \frac{Btu}{lb_{m}. °F}}{\frac{1 J}{kg.K}}) = 0.090 \frac{Btu}{lb_{m}. °F} $
Substituting the given and computed values:
$ ΔT = \frac{ΔQ}{mc_{p}}$
$ ΔT = \frac{ΔQ}{mc_{p}} = \frac{65 Btu}{(10 lb_{m}) (0.090 \frac{Btu}{lb_{m}.°F})} = 72.22 °F $
Finding $T_{f}$:
$T_{f} = T_{0} + ΔT = 77 °F + 72.22 °F = 149.22 °F $
