Can you please go over ##q = mcDeltaT##?

The specific heat capacity, or simply specific heat ##(C)## of a substance is the amount of heat energy required to raise the temperature of one gram of the substance by one degree Celsius. Heat energy is usually measured in Joules ##(“J”)## or calories ##(“cal”)##.
The variables in the equation ##q = mCDeltaT## mean the following:
##” let:”##
##q=”heat energy gained or lost by a substance”##
##m=”mass (grams)”##
##C=”specific heat”##
##DeltaT=”change in temperature”##
Note that ##DeltaT## is always calculated as ##”final temperature “-” initial temperature”##, not the other way around.
Therefore, you can look at that equation like this if it helps:
##”heat energy gained or lost by a substance”=(“mass”)(“specific heat”)(DeltaT)##
Example
How much heat energy is required to raise the temperature of ##”55.0g”## of water from ##”25.0″^@”C”## to ##”28.6″^@”C”##? The specific heat of water is ##”4.18″”J”##/##”g”^@”C”##. This is a very well known specific heat value and will frequently show up in specific heat questions.
Unknown:
##q## in Joules ##(“J”)##
Known/Given:
specific heat of water ##(C)## = ##”4.18″”J”##/##”g”^@”C”##
mass of water = ##”55.0g”##
##DeltaT## = ##”28.6″^@”C”-“25.0″^@”C”## = ##”3.6″^@”C”##
Equation:
##q =mcDeltaT##
Solution:
##q=”55.0g” xx “4.18””J”##/##”g”^@”C”xx”3.6″^@”C”##
##q## = ##”827.64″”J”##, which rounds to ##8.28xx10^2″J”## due to significant figures.

Calorimetry: Full lesson.

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