91Ó°ÊÓ

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

Let’s assume heat = q

q_copper = - q_water

We know that heat is related to mass, specific heat, and temperature as follows:

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

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

C_copper×m_copper× (Tf,_copper –T i,_copper) = C_water×m_water× (Tf,_water –T i,_water)

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

The initial temperature of copper is 314°C. The initial temperature of the water and the final temperature of copper is 22.6 °C.

As we know,

Specific heat of copper = 0.385 g/ °C

Specific heat of water = 4.184 g/ °C

The formula is:

\({{\rm{C}}_{{\rm{copper}}}}{\rm{ \times }}{{\rm{m}}_{{\rm{copper}}}}{\rm{ \times (Tf}}{{\rm{,}}_{{\rm{copper}}}}{\rm{--Ti}}{{\rm{,}}_{{\rm{copper}}}}{{\rm{)}}_{}}{\rm{\; = \; - }}{{\rm{C}}_{{\rm{water}}}}{\rm{ \times }}{{\rm{m}}_{{\rm{water}}}}{\rm{ \times (Tf}}{{\rm{,}}_{{\rm{water,copper}}}}{\rm{--Ti}}{{\rm{,}}_{{\rm{water, copper}}}}{\rm{)}}\)

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

\(\begin{array}{c}{\rm{0}}{\rm{.385 }} \times {\rm{ 248 }} \times {\rm{ (22}}{\rm{.6 ^\circ C\; - 314^\circ C) = - 4}}{\rm{.184 }} \times {\rm{390 }} \times {\rm{(Tf}}{{\rm{,}}_{{\rm{water,copper}}}}{\rm{ - 22}}{\rm{.6)}}\\{\rm{Tf}}{{\rm{,}}_{{\rm{water,copper}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.385 }} \times {\rm{ 248 }} \times {\rm{ (22}}{\rm{.6 ^\circ C\; - 314^\circ C)}}}}{{{\rm{ - 4}}{\rm{.184 }} \times {\rm{ 390}}}}{\rm{ + 22}}{\rm{.6}}\\{\rm{Tf}}{{\rm{,}}_{{\rm{water,copper}}}}{\rm{ = 38}}{\rm{.70 ^\circ C}}\end{array}\)