Question

How do you calculate specific heat without the final temp to calculate ΔT?

If you have a 100 g piece of iron at 100°C and place it into 500g of water at 20 °C will the temperature be above, below, or equal to 50°C?

Answer 1

You find it in your textbook...

`c_("Fe") = "0.450 J/g"^@ "C"`
`c_w = "4.184 J/g"^@ "C"`

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You don't need the final temperature explicitly. Suppose both masses were `"100 g"` and both objects were the same. Then they must reach halfway between `100^@ "C"` and `20^@ "C"`, which is `60^@ "C"`.

But iron has a lower specific heat capacity than water does, so it will not withstand as much heat as water would before changing temperature.

Thus, for the same masses of `"100 g"` each, `DeltaT_("Fe") > DeltaT_("H"_2"O")`, and the final temperature would be much closer to the starting temperature of water.

If we further have `"500 g"` of water instead of `"100 g"`, it will result in a temperature even closer to the starting temperature of water than with `"100 g"` of water.

Thus, `color(blue)(20^@ "C" < T_f < 50^@ "C")`, and it will be probably `20"-"`something `""^@ "C"`.

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If you want to actually calculate it... invoke conservation of energy:

`q_"Fe" + q_w = 0`

`q_"Fe" = -q_w`

Recall that `q = mcDeltaT`, with `q` being heat flow in `"J"`, `m` mass in `"g"`, `c` specific heat capacity in `"J/g"^@ "C"`, and `DeltaT` the change in temperature in `""^@ "C"`. Thus:

`m_"Fe"c_"Fe"DeltaT_"Fe" = -m_wc_wDeltaT_w`

`m_"Fe"c_"Fe"(T_f - T_"Fe") = -m_wc_w(T_f - T_w)`

Solve for the final temperature.

`m_"Fe"c_"Fe"T_f - m_"Fe"c_"Fe"T_"Fe" = -m_wc_wT_f + m_wc_wT_w`

`m_wc_wT_f + m_"Fe"c_"Fe"T_f = m_wc_wT_w + m_"Fe"c_"Fe"T_"Fe"`

The final temperature is then:

`color(blue)(T_f) = (m_wc_wT_w + m_"Fe"c_"Fe"T_"Fe")/(m_wc_w + m_"Fe"c_"Fe")`

`= ("500 g" cdot "4.184 J/g"^@ "C" cdot 20^@ "C" + "100 g" cdot "0.450 J/g"^@ "C" cdot 100^@ "C")/("500 g" cdot "4.184 J/g"^@ "C" + "100 g" cdot "0.450 J/g"^@ "C")`

`= color(blue)(21.7^@ "C")`