# Question d33b2

Mar 26, 2015

The original temperature of the spoon was $\text{5"^@"C}$.

First thing first, you need the specific heat values for copper and water (at that temperature).

${c}_{\text{copper") = 0.39"J"/("g" * ^@"C}}$

${c}_{\text{water at 90"^@"C") = 4.205"J"/("g" * ^@"C}}$

So, you've got your copper spoon and you place it in a volume of water.

The very first observation that you can make is that, since the temperature of the water drops, the initial temperature of the copper must have been lower than the equilibrium temperature, i.e. the copper got warmer (significantly warmer, I might add).

Since you're dealing with volume of water, which is not very useful in this case, use water's density to determine what mass of water you have

$\rho = \frac{m}{V} \implies m = \rho \cdot V$

m_("water at 90"^@"C") = 0.965"g"/cancel("mL") * 500cancel("mL") = "482.5 g"

Assuming no heat is lost to the outside of the system, the heat lost by the water will be equal to the heat gained by the copper (in absolute terms)

${q}_{\text{gained") = -q_("lost}}$

Since the relationship between heat and temperature change is given by

$q = m \cdot c \cdot \Delta T$, you'll have

${m}_{\text{copper") * c_("copper") * DeltaT_("copper") = -m_("water") * c_("water") * DeltaT_("water}}$

You know the temperature change for the water to be

DeltaT_("water") = T_("final") - T_("initial") = 88.55^@C - 90^@C = -"1.45"^@"C"

This means that you can write, after getting rid of the minus sign by using ${T}_{\text{initial") - T_("final}}$ for water

$100 \cancel{\text{g") * 0.39cancel("J")/(cancel("g") * ^@cancel("C")) * (88.55 - color(red)(T_("initial copper"))) = 482.5cancel("g") * 4.205cancel("J")/(cancel("g") * ^@cancel("C")) * 1.45^@cancel("C}}$

$3453.45 - 100 \cdot \textcolor{red}{{T}_{\text{initial copper}}} = 2941.9$, or

color(red)(T_("initial copper")) = 511.55/100 = "5.1155"^@"C"#

Rounded to one sig fig, the number of sig figs given for 100 g of copper and 500 mL of water, the answer will be

${T}_{\text{initial copper") = color(red)(5^@"C}}$

SIDE NOTE The significant difference between the drop in temperature of the water and the increase in temperature of the copper is due to the difference in their specific heat values.

A lot more energy needs to be removed in order to decrease the temperature of the water by 1 degree Celsius than it is needed to increase the temperature of the copper by the same amount.