Question #c2fdf

2 Answers
Mar 10, 2015

I would use the expression for heat exchanged #Q# as:
#Q=mCDeltaT#
#Q_(water)+Q_(iron)=0#

where:
#m# is mass
#C# heat capacity

#180*4.18(T_f-10)+1*0.48*(T_f-175)=0#
#752.4T_f-7524+0.48T_f-84=0#
#T_f=7608/752.88=10.1 °C#

Mar 12, 2015

The temperature at equalibirum, #T_(eq)#, which is also the final temperature, #T_(f)#, is #10^("o")"C"#.

This is a thermal equilibrium problem, in which objects of different temperatures that are in contact eventually reach the same temperature. The hotter object loses heat, while the colder object gains heat. According to the law of conservation of energy, the amount of heat lost should equal the amount of heat gained. In this situation, the final temperature is the equilibrium temperature, #"T"_("eq")#.

The equation for heat capacity is #"Q"# = #"mc"##Delta##"T"#, in which #"Q"# is heat energy in Joules, #"m"# is mass in grams, #"c"# is specific heat capacity in #"J/g"^("o")"C"#, and #Delta##"T"# is change in temperature (#T_("f")# - #T_("i")#).

For Iron:
#"m"# = #"1g"#
#"c"# = #"0.45 J/g"^("o")"C"^"[1]"#

#"T"_("i")"# #=# #"175"^("o")"C"#
#T_f = T_(eq) =?#

[1] Different sources report slight variations for the specific heat capacity of iron. The source I referenced is http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html

For Water:
#"m"# = #"180g"#
#"c"# = #"4.18 J/g"^("o")"C"#
#"T"_("i")"# #=# #"10"^("o")"C"#
#T_f = T_(eq) =?#

Solution:
We must set #-Q_(iron)# = #Q_(water)#. The heat capacity for iron is negative because the iron loses heat.

#"-(1)(0.45)"##(T_("eq") - 175)"# = #"(180)(4.18)"##(T_("eq") - 10)"#

#-.45(T_(eq) - 175)# = #752.4(T_(eq) - 10)"#

#-.45T_(eq) + 78.75# = #752.4T_(eq) - 7524#

#-.45T_(eq) + 78.75 + 7524# = #752.4T_(eq) - 7524 + 7524#

#-.45T_(eq) + 7602.75# = #752.4T_(eq)#

#-.45T_(eq) + 45T_(eq) + 7602.75# = #"752.4T_(eq) + 45T_(eq)#

#7602.75# = #797.4T_(eq)#

#(7602.75)/(797.4)# = #(797.4T_(eq))/(797.4)#

#"9.53"^("o")"C"# = #T_(eq)# = #T_(f)#

Round to #"10"^("o")"C"# due to the fact that the initial temperature of the water, #10^("o")"C"#, has only one significant figure. Basically, the 1g piece of heated iron had a negligible impact on the temperature of the 180g mass of cold water.

Reference: http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Chapter12.htm