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How much heat is released when 100.0 mL of water is called from 95.0°C to 0°C ?

Answer 1

#~~4.0xx10^4# #"J"# are released.

The equation to use is:

#q=mcDeltaT#,

where:

#q# is heat energy, #m# is mass, #c# is specific heat capacity, and #DeltaT# is change in temperature. #DeltaT=T_"final"-T_"initial"#

The density of water is #"1 g/mL"#, which means that every #"mL"# has a mass of #"1 g"#. Therefore, #"100.0 mL"# of water has a mass of #"100.0 g"#.

Since there is no mention of a phase change, I'm assuming that the water is still liquid at #0^@"C"#.

Known

#m="100.0 g"#

#c_"H2O"="4.184 J/g"*""^@"C"#

#T_"initial"="95.0"^@"C"#

#T_"final"="0"^@"C"#

#DeltaT="95.0"^@"C"-0^@"C"="95"^@"C"#

Unknown

#q#

Solution

Plug in the known values and solve.

#q=100.0color(red)cancel(color(black)(color(red)cancel(color(black)("g"))))xx(4.184"J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C")))xx95^@color(red)cancel(color(black)("C"))=4.0xx10^4# #"J"# #("39748 J"# rounded to two significant figures)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.