Calculate the heat absorbed by 125 g of water (SH = 1.00 cal/g°C) when the temperature rises from 21.0°C to 35.0°C?

1 Answer
zhirou
Jan 29, 2018

#"1750cal"#, #7320J#, or #7.32kJ#

Explanation:

In this question, we'll need to use this equation:

#q=mCDeltaT#

where #q# is the heat energy gained or lost by a substance
#m# is the mass of the substance, #C# the specific heat of that substance,
and #DeltaT# the change in temperature (celsius).

For a more thorough explanation, check out this great answer!

In our case, we need to calculate #q# given #"125g"#, a specific heat of #"1 cal/g°C"#, and the initial and final temperature of #21.0°C# and #35.0°C#.
#DeltaT# is #"final temperature - initial temperature"#, so it would be #"35°C - 21°C = 14°C"#.

Now, just plug everything in and find #q#:
#q=mcDeltaT#
#q="125g ⋅ 1 cal/g°C ⋅ 14°C"#
#q="1750cal"#

If we wanted to convert that to Joules, which is the SI unit for energy, we would multiply it by #4.184#.

#q="1750cal" = (1750*4.184)J = 7322J#

Since #3# significant figures are given in the question, we should only have #3# significant figures in the answer. That makes it #7320J#.

We could also convert this to kilojoules by dividing it by #1000#.
#(7320J)/1000=7.32kJ#

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