Question #d4a35

1 Answer
Apr 14, 2016

Answer:

#"12.8 kJ"#

Explanation:

The key to this problem lies with the value of water's specific heat, which is said to be equal to

#c_w = "4.184 J g"^(-1)""^@"C"^(-1)#

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

In your case, water's specific heat tells you that you need #"4.184 J"# of energy in the form of heat in order to increase the temperature of #"1 g"# of water by #1^@"C"#.

Notice that specific heat is expressed in joules per gram per degree Celsius. This means that you can break up the calculation into two parts

  • determine how much heat you need to add in order to increase the temperature of #"45.0 g"# of water by #1^@"C"#

  • use this value to determine how much heat would be needed in order to increase the sample's temperature by #DeltaT#

In this case, #DeltaT#, which represents change in temperature, will be equal to

#DeltaT = T_"final" - T_"initial"#

#DeltaT = 74.0^@"C" - 6.2^@"C" = 67.8^@"C"#

So, to increase the temperature of #"45.0 g"# of water by #1^@"C"#, you'd need

#45.0 color(red)(cancel(color(black)("g"))) * overbrace(("4.184 J"^@"C"^(-1))/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat needed per 1"^@"C")) = "188.28 J"^@"C"^(-1)#

If this is how much heat you need to produce a #1^@"C"# increase in temperature for a #"45.0-g"# sample of water, it follows that a #67.8^@"C"# increase in temperature would require

#67.8 color(red)(cancel(color(black)(""^@"C"))) * "188.28 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "12,765.4 J"#

I'll leave the answer rounded to three sig figs and express it in kilojoules

#"heat needed" = color(green)(|bar(ul(color(white)(a/a)"12.8 kJ"color(white)(a/a)|)))#

ALTERNATIVE APPROACH

You can also solve this problem by using the following equation

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#, where

#q# - the amount of heat gained / lost
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as the difference between the final temperature and the initial temperature

Plug in your values to get

#q = 45.0 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 67.8color(red)(cancel(color(black)(""^@"C")))#

#q = color(green)(|bar(ul(color(white)(a/a)"12.8 kJ"color(white)(a/a)|)))#