Answer:
6700.6 BTU
Step-by-step explanation:
First you have to warm the block from -15° F to 32°F, the heat to needed to do this is:
[tex]Q_{1}=mc\Delta T=20lb*0.49\frac{BTU}{lbF} *(32F-(-15F))=460.6BTU[/tex]
After you need to melt the ice. The heat you need is:
[tex]Q_{2}=mL=20lb*144\frac{BTU}{lb}=2880BTU[/tex]
Finally, you need to heat the water from 32° F to 200 ° F, and the heat for this is:
[tex]Q_{3}=mc\Delta T=20lb*1\frac{BTU}{lbF} *(200F-32F)=3360BTU[/tex]
[tex]Q=Q_{1}+Q_{2}+Q_{3}=460.6BTU+2880BTU+3360BTU=6700.6BTU[/tex]To have the total heat you used you have to su[tex]Q_{1},Q_{2}[/tex] and [tex]Q_{3}[/tex]
