Answer:
[tex]1.21\times 10^4[/tex] cal would be needed to heat 5.0 lbs of copper from 22 degrees C to 80.0 degrees C.
Explanation:
[tex]Q=m \times c \times \Delta T[/tex]
where
[tex]\Delta T[/tex] = Final T - Initial T
Q is the heat energy in calories
c is the specific heat capacity (for copper 0.092 cal/(g℃))
m is the mass of water
plugging in the values
[tex]$Q=5.01 b s \times 0.092 \frac{c a l}{g^{\circ} \mathrm{c}} \times\left(80.0^{\circ} \mathrm{C}-22^{\circ} \mathrm{C}\right)$[/tex]
Please Note:
1 lb = 453.592grams
So,
5 lbs = 5 × 453.592g = 2268 g
[tex]$\begin{aligned} Q &=2268 g \times 0.092 \frac{c a l}{g^{\circ} \mathrm{C}} \times 58^{\circ} \mathrm{C} \\\\ Q &=12102 \mathrm{cal} \end{aligned}$[/tex]
[tex]=1.21\times10^4[/tex] cal (Answer)
