Answer:
cal would be needed to heat 5.0 lbs of copper from 22 degrees C to 80.0 degrees C.
Step-by-step explanation:
![Q=m * c * \Delta T](https://img.qammunity.org/2020/formulas/chemistry/middle-school/ouf3u6o25lt5mfy3e3w7opgdymnufnah4n.png)
where
= 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
![$Q=5.01 b s * 0.092 \frac{c a l}{g^(\circ) \mathrm{c}} *\left(80.0^(\circ) \mathrm{C}-22^(\circ) \mathrm{C}\right)$](https://img.qammunity.org/2020/formulas/chemistry/middle-school/81i9eztmz89tdvczs52qchz1a7avo81hyv.png)
Please Note:
1 lb = 453.592grams
So,
5 lbs = 5 × 453.592g = 2268 g
![$\begin{aligned} Q &=2268 g * 0.092 \frac{c a l}{g^(\circ) \mathrm{C}} * 58^(\circ) \mathrm{C} \\\\ Q &=12102 \mathrm{cal} \end{aligned}$](https://img.qammunity.org/2020/formulas/chemistry/middle-school/twe13a2kmsqpz7dz49g6ktliwsk8g31ce3.png)
cal (Answer)