Question

A 248-g piece of copper is dropped into 390 mL of water at 22.6°C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.

Short answer

The initial temperature of the piece of copper = 255.75°C.

Step-by-step solution

Given data

• m_copper = 248 g.

• Copperis dipped into water at 22.6 °C.

• The initial temperature of the water is 22.6 °C.

Heat transfer between copper and water

The heat transfer is entirely between copper and water and there is no loss of heat to the surroundings. So,

Heat given off by copper = - Heat absorbed by water

q_copper = - q_water

As we know that heat is related to mass, specific heat, and temperature which is given as follows.

\({{\rm{(c \times m \times \Delta T)}}_{{\rm{Copper}}}}{\rm{\; = \;\;(c \times m \times \Delta T}}{{\rm{)}}_{{\rm{Water}}}}\)

Let ‘f’ be the final and ‘i’ the initial terms and On expanding the above equation, the new equation becomes:

\({{\rm{C}}_{{\rm{copper}}}}{\rm{ \times }}{{\rm{m}}_{{\rm{copper}}}}{\rm{ \times (}}{{\rm{T}}_{\rm{f}}}_{_{{\rm{copper}}}}{\rm{--}}{{\rm{T}}_{\rm{i}}}_{_{{\rm{copper}}}}{{\rm{)}}_{}}{\rm{\; = \;\;}}{{\rm{C}}_{{\rm{water}}}}{\rm{ \times }}{{\rm{m}}_{{\rm{water}}}}{\rm{ \times (T}}{{\rm{f}}_{{\rm{water}}}}{\rm{--T}}{{\rm{i}}_{{\rm{water}}}}{\rm{)}}\)

Density of water is 1.0 g/mL; so 390 mL water = 390 g water.

The initial temperature of the water is 22.6°C and the final temperature of water and copper is 39.9 °C.

Calculation of initial temperature of copper

As we know,

• Specific heat of copper = 0.385 g/°C.
• Specific heat of water = 4.184 g/ °C.

The formula is:

\({{\rm{C}}_{{\rm{copper}}}}{\rm{ \times }}{{\rm{m}}_{{\rm{copper}}}}{\rm{ \times (}}{{\rm{T}}_{\rm{f}}}_{_{{\rm{copper}}}}{\rm{--}}{{\rm{T}}_{\rm{i}}}_{_{{\rm{copper}}}}{{\rm{)}}_{}}{\rm{\; = \;\;}}{{\rm{C}}_{{\rm{water}}}}{\rm{ \times }}{{\rm{m}}_{{\rm{water}}}}{\rm{ \times (T}}{{\rm{f}}_{{\rm{water}}}}{\rm{--T}}{{\rm{i}}_{{\rm{water}}}}{\rm{)}}\)

By putting the given values into the above equation, we get,

\(\begin{array}{c}{\rm{0}}{\rm{.385 \times 248 \times (39}}{\rm{.9 ^\circ C\; - Ti}}{{\rm{,}}_{{\rm{copper}}}}{\rm{) = - 4}}{\rm{.184 \times 390 \times (39}}{\rm{.9-- 22}}{\rm{.6)}}\\{\rm{0}}{\rm{.385 \times 248 \times (39}}{\rm{.9 ^\circ C\; - Ti}}{{\rm{,}}_{{\rm{copper}}}}{\rm{)\;\;\; = \;\;1631}}{\rm{.76 \times 17}}{\rm{.3}}\\{\rm{39}}{\rm{.9 ^\circ C\; - Ti}}{{\rm{,}}_{{\rm{copper}}}}{\rm{\;\; = }}\frac{{{\rm{1631}}{\rm{.76 \times 17}}{\rm{.3}}}}{{{\rm{0}}{\rm{.385 \times 248}}}}\\{\rm{ - Ti}}{{\rm{,}}_{{\rm{copper}}}}{\rm{ = }}\left( {{\rm{295}}{\rm{.65 -- 39}}{\rm{.9}}} \right){\rm{^\circ C }}\\{\rm{ = 255}}{\rm{.75^\circ C\;}}\end{array}\)