Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

# 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.

SMARTERTEACHER YouTube

Calorimetry: Full lesson.