Question

Question #c1c11

Answer 1

`17.1 ^@"C"`

Explanation:

We can use the formula `q = m * c * DeltaT`, where `q` is the heat content, `m` is the mass in grams, `c` is the specific heat, and `DeltaT` is the change in temperature `(T_f - T_i)`.

The zinc has a greater heat content than the water, so when the zinc is placed into the water, it will lose heat (i.e., its temperature will decrease) and the water will gain heat (its temperature will increase). The final temperature should thus be somewhere in between `12.5^@"C"` and `67^@"C"`.

The final temperature of both will be the same, so the heat lost by the zinc will equal the heat gained by the water.

`-q_"Zn" = q_("H"_2"O")`

`-(m * c * DeltaT)_("Zn") = (m * c * DeltaT)_("H"_2"O")`

You can look up the specific heats of the zinc and water.

`-("17800 g" * ("0.39 J")/("g" * °"C") * (T_f - 67)°"C")_("Zn") = ("18000 g" * ("4.18 J")/("g" * °"C") * (T_f - 12.5)°"C")_("H"_2"O")`

Now the units cancel out.

`-("17800" cancel"g" * ("0.39 J")/(cancel"g" * cancel(°"C")) * (T_f - 67)cancel(°"C")) = "18000" cancel"g" * ("4.18 J")/(cancel"g" * cancel(°"C")) * (T_f - 12.5)cancel(°"C")`

`-("6942 J" * (T_f - 67)) = "75240 J" * (T_f - 12.5)`

`-("6942 T"_f - 465114 ) = "75240 T"_f - 940500`

`-"6942 T"_f + 465114 = "75240 T"_f - 940500`

`-"82182 T"_f = -1405614`

`T_f = 17.1 ^@"C"`

Answer 2

The final temperature of the water is 17.1 °C.

Explanation:

The guiding principle is the Law of Conservation of Energy: the sum of all the energy changes must add up to zero.

The formula for the heat `q` gained or lost by a substance is

`color(blue)(bar(ul(|color(white)(a/a)q = mCΔTcolor(white)(a/a)|)))" "`

where

`m` is the mass of the substance.
`C` is its specific heat capacity.
`ΔT = T_"f" - T_"i"` is the change in temperature.

In this problem, there are two heat flows.

`"Heat lost by zinc + Heat gained by water" = 0`

`color(white)(mmmmm)q_1 color(white)(mmmll)+color(white)(mmmmm) q_2color(white)(mmmml) = 0`

`color(white)(mmm)m_1c_1ΔT_1 color(white)(mml)+color(white)(mmm) m_2c_2ΔT_2color(white)(mmll) = 0`

The final temperature of the zinc will be the same as the final temperature of the water.

In this problem,

`m_1 = "17.8 kg";color(white)(mmmmmll) m_2 = "18 kg"`

`C_1 = "0.387 kJ·°C"^"-1""kg"^"-1"; color(white)(m) C_2 = "4.187 kJ·°C"^"-1""kg"^"-1"`

`ΔT_1 = T_"f"color(white)(l) "- 67 °C"; color(white)(mmm)ΔT_2 = T_"f" color(white)(l)"- 12.5 °C"`

`m_1C_1ΔT_1 + m_2C_2ΔT_2 = 0`

`17.8 color(red)(cancel(color(black)("kg"))) × 0.387 color(red)(cancel(color(black)("kJ·°C"^"-1""kg"^"-1"))) × (T_"f"color(white)(l) "- 67 °C")`
`+ 18 color(red)(cancel(color(black)("kg"))) ×4.187 color(red)(cancel(color(black)("kJ·°C"^"-1""kg"^"-1"))) × (T_"f" color(white)(l)"- 12.5 °C") =0`

`6.889( (T_"f"color(white)(l) "- 67 °C") + 75.37(T_"f" color(white)(l)"- 12.5 °C") = 0`

`6.889T_"f" - "461.5 °C" + 75.37T_"f" - "942.1 °C" =0`

`82.25 T_"f" = "1404 °C"`

`T_"f" = "1404 °C"/82.25 = "17.1 °C"`