Question #9a262

2 Answers
Jan 28, 2015

The symbol #"q"# is used to express either heat absorbed or heat lost, although heat lost is usually written as #"-q"#.

As you know, the formula for calculating heat gained is

#"q" = m * c * DeltaT"#, where

#m# - the mass of the substance;
#c# - its specific heat;
#DeltaT# - the difference between the final and the initial temperature of the substance.

In your case, you have a metal that is probably heated and then placed in water. The water would heat up by a number of degrees, while the metal would cool down by a much greater number of degrees. The heat lost by the metal will be equal to

#"-q" = m_("metal") * c_("metal") * DeltaT_("metal")#

This will get you

#c_("metal") = ("-q")/("m"_("metal") * DeltaT_("metal")#

Since the metal is cooling off, its final temperature will be lower than its initial one, so #DeltaT_("metal")# will be negative #-># the negative signs will cancel out.

As a conclusion, #"q"# represents either the heat lost by the hot metal, or the heat gained by the water, since what is lost by the metal must be gained by the water.

#-q_("metal") = q_("water")#

Jan 28, 2015

The equation for heat and temperature change is:

#Q=c*m*DeltaT#

Where #Q# = total warmth energy (in #J# for Joules)
#c# = specific heat capacity (in #J//g*K#)
#m# = mass (in #g# for grams)
#DeltaT# = temperature difference (after-before) (in #K# or #C#)

If #m# (mass) is twice as high, you will need twice #Q# for the same effect. Same goes for a greater #DeltaT# temperature difference.

If you want to determine #c# the equation becomes:

#c=Q/(m*DeltaT#

And it then depends how you supplied the heat. If you do this electrically (by a heating element) the other half of the equation becomes:

#E=U*I*t#

Where #E# = energie provided (also in #J#)
#U# = voltage (in #V#)
#I# = current (in #A#)
#t# = time (in #s# seconds)

Assuming there are no losses (but there always are), you can set #E# equal to #Q# to get:

#c=(U*I*t)/(m*DeltaT#