Question

A six pack of your favorite beverage (12 fl. oz. cans, assumed to have the same characteristics as water) is placed in a cooler at 70° F. How much heat must be removed (BTU) to cool the beverages to 34°F?

Answer 1

Answer:

$Q = 169 BTU$

Explanation:

As we know that volume is given as

$V = 12 Fl oz$

so it is given in liter as

$V = 12 fl oz = 0.355 Ltr$

now we have six pack of such volume

so total volume is given as

$V = 6\times 0.355 Ltr$

$V = 2.13 ltr$

so its mass is given as

$m = 2.13 kg$

now the change in temperature is given as

$\Delta T = 70 - 34 = 36 ^oF$

$\Delta T = 20^oC$

now the heat given to the liquid is given as

$Q = ms\Delta T$

$Q = 2.13(4186)(20)$

$Q = 1.78 \times 10^5 J$

$Q = 169 BTU$