A 100-gram sample of #H_2O(l)# at 22.0°C absorbs 8360 joules of heat. What will be the final temperature of the wild?

1 Answer
Truong-Son N.
Jan 23, 2018

Well, by conservation of energy, the WILD reaches #42.0^@ "C"#. By the way, you can't find the original temperature of the WILD, unless the WILD has a known mass and specific heat capacity.

Well, heat only flows until one reaches a so-called thermal equilibrium, where there isn't a temperature difference.

#q_w = m_wC_PDeltaT_w#

where #q# is the heat flow, #m# is the mass in #"g"#, #C_P# is the specific heat capacity in #"J/g"^@ "C"#, and #DeltaT# is the change in temperature.

In absorbing #"8360 J"# of heat, #q_w > 0#. Thus,

#q_w = "8360 J" = "100.0 g" cdot "4.184 J/g"^@ "C" cdot (T_f - 22.0^@ "C")#

As a result, the water must become hotter.

#color(blue)(T_f) = ("8360 J")/("100.0 g" cdot "4.184 J/g"^@ "C") + 22.0^@ "C"#

#= color(blue)(42.0^@ "C")#

The surroundings then lose heat in such a way that they cool down from above #42.0^@ "C"# to reach #42.0^@ "C"#.

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