If 65.34g piece of tin were to lose 754 Joules of energy in a calorimeter while experiencing a temperature change from 23.9 degrees C to 22.0 degrees C, how would you find its specific heat?

1 Answer
Oct 27, 2015

Answer:

The values you provided are way off.

Explanation:

SIDE NOTE First thing first, the values you provided are very inaccurate. Using these values will produce an impossible result for the specific heat of tin, approximately #30# times bigger than what its real value is.

WIth this being said, I will assume that the piece of tin lost #"75.4 J"#, instead of #"754 J"#. The result will not be ideal, but it will be closer to reality.

#stackrel("-----------------------------------------------------------------------------------------------------------------------")#

The question wants you to determine tin's specific heat.

Before doing any calculation, try to get a clear understanding of what you need to determine.

A substance's specific heat tells you the amount of heat needed to increase the temperature of #"1 g"# of that substance by #1^@"C"#.

In your case, a mroe accurate description would sound like this - the amount of heat that must be lost by #"1 g"# of a substance to decrease its temperature by #1^@"C"#.

The equation that establishes a relationship between heat lost and change in temperature looks like this

#color(blue)(-q = m * c * DeltaT)" "#, where

#-q# - heat lost by the metal
#m# - the mass of the metal
#c# - its specific heat
#DeltaT# - th change in temperature, defined as final temperature minus initial temperature

The trick here is to realize that when heat is being lost by the system, #q# will have a negative value. In this case, #"75.4 J"# are being lost, so #q# will be

#q = -"75.4 J"#

Since you have all the information you need to find the specific heat of tin, plug in your values in the above equation and solve for #c#

#c = (-q)/(m * DeltaT)#

#c = (-(-"75.4 J"))/("65.34 g" * (22.0 - 23.9)^@"C") = color(green)(0.607"J"/("g" ""^@"C"))#

The actual specific heat of tin is #0.21"J"/("g" ""^@"C")#, so I recommend double-checking the values you have.