Question #087c4

1 Answer
May 14, 2017

Answer:

#91.6^@"C"#

Explanation:

For starters, you need to make a note that the specific heat of water is equal to

#c_"water" = "1 cal J g"^(-1)""^@"C"^(-1)#

This means that when the temperature of #"1 g"# of liquid water decreases by #1^@"C"#, #"1 cal"# of heat is being given off.

Now, your tool of choice here will be this equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

  • #q# is the heat gained by the water
  • #m# is the mass of the water
  • #c# is the specific heat of water
  • #DeltaT# is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

You need to solve for #DeltaT#, so rearrange the equation as

#DeltaT = q/(m * c)#

Plug in your values to find

#DeltaT = (1794 color(red)(cancel(color(black)("cal"))))/(250 color(red)(cancel(color(black)("g"))) * 1 color(red)(cancel(color(black)("cal"))) color(red)(cancel(color(black)("g"^(-1)))) ""^@"C"^(-1)) = 7.2^@"C"#

Now, you know that the sample is losing heat, so you can say that the final temperature will be #7.2^@"C"# lower than the initial temperature

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

which, in your case, will be equal to

#T_"final" = 98.8^@"C" - 7.2^@"C" = color(darkgreen)(ul(color(black)(91.6^@"C")))#

The answer is rounded to one decimal place.

#color(white)(a)#

SIDE NOTE It's important to notice that if you use

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

you get

#T_"final" = DeltaT + T_"initial"#

#color(red)(cancel(color(black)(T_"final" = 7.2^@"C" + 98.8^@"C" = 106^@"C"))) -># not good

The problem here is that the sample is actually giving off heat, so you should use

#q = - "1794 cal"#

in the equation. The minus sign is needed because the final temperature is lower than the initial temperature. This will get you

#DeltaT = (-1794 color(red)(cancel(color(black)("cal"))))/(250 color(red)(cancel(color(black)("g"))) * 1 color(red)(cancel(color(black)("cal"))) color(red)(cancel(color(black)("g"^(-1)))) ""^@"C"^(-1)) = -7.2^@"C"#

which results in

#T_"final" = - 7.2^@"C" + 98.8^@"C" = 91.6^@"C" -># good