The specific heat capacity of liquid water is 4.18 kJ/g C, how would you calculate the quantity of energy required to heat 1.00 g of water from 26.5 C to 83.7 C?

1 Answer
Oct 26, 2015

Answer:

#"239 J"#

Explanation:

First thing first, you mistyped the specific heat of water, which should be

#c_"water" = 4.18"J"/("g" ""^@"C")#

Now, a substance's specific heat tells you how much heat is required to increase the temperature of #"1 g"# of that substance by #1^@"C"#.

In the case of water, you would need #"4.18 J"# to increase the temperature of #"1 g"# of water by #1^@"C"#.

Notice that your sample of water has a mass of #"1 g"# as well, which means that the only factor that will determine the amount of heat needed will be the difference in temperature.

The equation that establishes a relationshop between heat and change in temperature looks like this

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - heat absorbed
#c# - the specific heat of the substance, in your case of water
#DeltaT# - the change in temperature, defined as the difference between the final temperature and the initial temperature

Plug in your values and solve for #q# to get

#q = 1.00color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (83.7 - 26.5)color(red)(cancel(color(black)(""^@"C")))#

#q = "239.096 J"#

Rounded to three sig figs, the answer will be

#q = color(green)("239 J")#

SIDE NOTE Notice how much heat would be needed to increase the temperature of 57.2 g of water by #1^@"C"#. #DeltaT# will now be equal to #1^@"C"# and you'd have

#q = 57.2color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 1color(red)(cancel(color(black)(""^@"C")))#

#q = "239 J"#

This tells you that you need the same amount of heat to increase the temperature of #"1 g"# of water by #57.2^@"C"#, as you do to increase the temperature of #"57.2 g"# of water by #1^@"C"#.