If 200 grams of water is to be heated from 24°C to 100°C to make a cup of tea, how much heat must be added?

The specific heat of water is 4.18 #J##/##g*C#.

1 Answer
Dec 23, 2015

Answer:

#q = "64 kJ"#

Explanation:

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

In water's case, you know that its specific heat is equal to #4.18"J"/("g" ""^@"C")#. This tells you that in order to increase the temperature of #"1 g"# of water by #1^@"C"#, you need to supply #"4.18 J"# of heat.

How much heat would you need to increase the temperature of #"200 g"# of water by #1^@"C"#?

Well, if you need #"4.18 J"# per gram to increase its temperature by #1^@"C"#, it follows that you will need #200# times more heat to get this done.

Likewise, if you were to increase the temperature of #"1 g"# of water by #76^@"C"#, you'd need #76# times more heat than when increasing the temperature of #"1 g"# by #1^@"C"#.

Combine these two requirements and you get the total amount of heat required to increase the temperature of #"200 g"# of water by #76^@"C"#.

Mathematically, this is expressed using the following equation

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

#q# - heat absorbed/lost
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

#q = 200 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (100 - 24)color(red)(cancel(color(black)(""^@"C")))#

#q = "63536 J"#

I'll leave the answer rounded to two sig figs and expressed in kilojoules

#q = color(green)("64 kJ")#