Question

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?

Answer 1

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.

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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"`.