How many joules are required to heat 250 grams of liquid water from 0°C to 100°C?

1 Answer
May 28, 2016

Answer:

#Q= 104525 J#

Explanation:

The key equation is

#Q=mc Delta T#,

where #Q# is the heat energy required to heat a mass, #m#,
though a change in temperature of #Delta T # and #c# is the heat capacity of the material (found in books!). Typical units are #Q# in Joules or #J#, #m# in #kg# and #Delta T# is in either Kelvin, #K#, or degrees Celsius #""^oC#. The units of #c# depend on the other units used but #J // kg# #""^o C# is typical.

we look up #c# for water (textbook, the internet, etc)
#c=4181# # J// kg# #""^o C#
#m=250# #g=0.250# #kg#
#Delta T= T_"final" -T_"initial"=100# #""^oC-0# #""^oC=100# #""^oC#

#Q=(0.250# #kg)(4181 J // kg# #""^o C) (100# #""^o C)#
#Q=(0.250# #cancel{kg})(4181 J // cancel{kg}# #cancel{""^o C}) (100# #cancel{""^oC})#

#Q= 104525 J#