# Question f5994

May 12, 2015

The equilibrium temperature will be equal to ${70}^{\circ} \text{C}$.

In order to solve this problem, you need to know the specific heat of water, of copper, and of glass.

${c}_{\text{water" = "4.18 J/g"^@"C}}$
${c}_{\text{copper" = "0.386 J/g"^@"C}}$
${c}_{\text{glass" = "0.84 J/g"^@"C}}$

So, the idea is that the heat lost by the copper will be absorbed by the water and the glass. The equilibrium temperature will thus be lower than ${940}^{\circ} \text{C}$, and higher than ${20}^{\circ} \text{C}$.

Mathematically, this is written as

q_"copper" = -(q_"water" + q_"glass")

${q}_{\text{copper" = -q_"water" - q_"glass}}$ $\textcolor{b l u e}{\left(1\right)}$

The relationship between heat lost/gained and temperature change for a substance is given by

$q = m \cdot c \cdot \Delta T$, where

$q$ - heat;
$m$ - the mass of the substance;
$c$ - its specific heat;
$\Delta T$ - the change in temperature, defined as the difference between the equilibrium temperature and the initial temperature;

So, plug your data into equation $\textcolor{b l u e}{\left(1\right)}$ to get

m_"copper" * c_"copper" * (T_"ech" - T_"copper"^0) = -m_"water" * c_"water" * (T_"ech" - T_"water"^0) - m_"glass" * c_"glass" * (T_"ech" - T_"glass"^0)

$1000 \cancel{\text{g") * 0.386cancel("J")/(cancel("g") * ^@cancel("C")) * (T_"ech" - 940)^@cancel("C") = 1500cancel("g") * 4.18cancel("J")/(cancel("g") * ^@cancel("C")) * (T_"ech" - 20)^@cancel("C") - 500cancel("g") * 0.84cancel("J")/(cancel("g") ^@cancel("C")) * (T_"ech" - 20)^@cancel("C}}$

$386 \cdot {T}_{\text{ech" - 362840 = -6270 * T_"ech" + 125400 - 720 * T_"ech}} + 8400$

$7076 \cdot {T}_{\text{ech" = 496640 => T_"ech" = 496640/7076 = 70.19^@"C}}$

Rounded to one sig fig, the number of sig figs you gave for the mass of copper, the answer will be

T_"ech" = color(green)(70^@"C")#