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

Question

Jeremiah Adams · Accepted Answer

#Q= 104525 J#

The crucial formula 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#

Cameron Andrews · Answer

undefined The specific heat capacity of water is 4.18 J/g°C. The formula to calculate the energy required to heat a substance is Q = mcΔT, where Q is the energy in joules, m is the mass in grams, c is the specific heat capacity in J/g°C, and ΔT is the change in temperature in degrees Celsius. Substituting the values, we get: Q = (250 g) * (4.18 J/g°C) * (100°C - 0°C) = 104500 J. Therefore, 104500 joules are required to heat 250 grams of liquid water from 0°C to 100°C.