Answer:
6700.6 BTU
Explanation:
First you have to warm the block from -15° F to 32°F, the heat to needed to do this is:
![Q_(1)=mc\Delta T=20lb*0.49(BTU)/(lbF) *(32F-(-15F))=460.6BTU](https://img.qammunity.org/2020/formulas/mathematics/college/t6ree55v49834wyvwti43psy22cc03i5ad.png)
After you need to melt the ice. The heat you need is:
![Q_(2)=mL=20lb*144(BTU)/(lb)=2880BTU](https://img.qammunity.org/2020/formulas/mathematics/college/71560ynb91chipids2i94eyefn4lbny9tu.png)
Finally, you need to heat the water from 32° F to 200 ° F, and the heat for this is:
![Q_(3)=mc\Delta T=20lb*1(BTU)/(lbF) *(200F-32F)=3360BTU](https://img.qammunity.org/2020/formulas/mathematics/college/yn7iyonxrdln5mnje5p24srajcap0f9cdn.png)
To have the total heat you used you have to su
and