Question

A 1 kg bar of gold at 150°C was placed in 3 kg of water that was originally at 20°C. What happened to the water?

Answer 1

The temperature of the water rose by about 1 °C.

Assume that the gold was originally at 20 °C.

We can calculate the heat that was added to heat it to 150 °C by using the specific heat formula:

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

`m = "1000 g"`
`c = "0.126 J·°C·g"^"-1"`
`ΔT = "(150 - 20) °C" = "130 °C"`

∴ `q = 1000 color(red)(cancel(color(black)("g"))) × 0.126 "J"·color(red)(cancel(color(black)("°C·g"^"-1"))) × 130 color(red)(cancel(color(black)("°C"))) = "16 380 J"`

This heat will be transferred to the water, warming it from 20 °C to some higher temperature.

Calculating the new temperature of the water

We can use the same formula as before:

`q = "16 380 J"`
`m = "3000 g"`
`c = "4.184 J·°C"^"-1""mol"^"-1"`
`ΔT = T_2 - T_1 = T_2 - "20 °C"`

`q = mcΔT`

`"16 380" color(red)(cancel(color(black)("J"))) = 3000 color(red)(cancel(color(black)("g"))) × 4.184 color(red)(cancel(color(black)("J")))·"°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × (T_2 "- 20 °C") = "12 552"T_2 "°C"^"-1" - "251 040"`

`"16 380" = "12 552T"_2 "°C"^"-1" - "251 040"`

`T_2 = "251 040 + 16 380"/("12 552°C"^"-1") = "267 420"/"12 522" "°C" = "21.4 °C"`

You may be surprised that the temperature of the water increases by only 1.4 °C.

Remember, though, that the specific heat capacity of water is 33 times that of gold, and the mass of water is 3 times the mass of gold.

That makes a factor of 100.

Hence, `ΔT` for the water is only `1/100` the `ΔT` for the gold.