# 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?

Jun 15, 2016

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)|)))" "

$m = \text{1000 g}$
$c = \text{0.126 J·°C·g"^"-1}$
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:

$q = \text{16 380 J}$
$m = \text{3000 g}$
$c = \text{4.184 J·°C"^"-1""mol"^"-1}$
ΔT = T_2 - T_1 = T_2 - "20 °C"

q = mcΔT

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

$\text{16 380" = "12 552T"_2 "°C"^"-1" - "251 040}$

${T}_{2} = \text{251 040 + 16 380"/("12 552°C"^"-1") = "267 420"/"12 522" "°C" = "21.4 °C}$

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 $\frac{1}{100}$ the ΔT for the gold.