Question

A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to convert the ice to water at 200'F?

Answer 1

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:

$Q_{1}=mc\Delta T=20lb*0.49\frac{BTU}{lbF} *(32F-(-15F))=460.6BTU$

After you need to melt the ice. The heat you need is:

$Q_{2}=mL=20lb*144\frac{BTU}{lb}=2880BTU$

Finally, you need to heat the water from 32° F to 200 ° F, and the heat for this is:

$Q_{3}=mc\Delta T=20lb*1\frac{BTU}{lbF} *(200F-32F)=3360BTU$

$Q=Q_{1}+Q_{2}+Q_{3}=460.6BTU+2880BTU+3360BTU=6700.6BTU$To have the total heat you used you have to su$Q_{1},Q_{2}$ and $Q_{3}$