Physics & Chemistry Question...

Question

Dear experts,

Im having trouble answering the following question:

How many calories (cal) does it take to bring 80

liters of water from 20 ° C to 80 ° C?

0 4800 kcal

0 1.5 kcal

0 1600 cal

0 3200 kcal

0 6400 cal

The only thing i know is that 1cal = 4.18 J (I don't know how to solve this)...

Answer 1

You can use the formula q = m∆T

q = heat = ?

m = mass of water = 80 liters = 80,000 mls = 80,000 g (assuming a density of 1g/ml for water)

C = specific heat of water = 4.184 J/gº = 1 cal/gº

∆T = change in temperature = 80º - 20º = 60º

q = (80,000 g)(1 cal/gº)(60º)

q = 4,800,000 cal = 4800 kcal