91Ó°ÊÓ

Under ideal circumstances, the net heat change of a reaction is zero, \({q_{subs\tan ceM}} + {q_{subs\tan ceN}}\) = 0.

This relationship can be rearranged to show that the heat gained by a substance M is equal to the heat lost by a substance W; \({q_{subs\tan ceM}} = - {q_{subs\tan ceN}}\).

The negative sign merely shows that the direction of the heat flow is opposite to each other.

Let us assume that substance M is coffee and substance N is silver.

\({{\bf{Q}}_{\bf{M}}}{\bf{ = }}{{\bf{C}}_{\bf{M}}}{\bf{ \times }}{{\bf{m}}_{\bf{M}}}{\bf{ \times \Delta }}{{\bf{T}}_{\bf{M}}}\)and \({{\bf{Q}}_{\bf{N}}}{\bf{ = }}{{\bf{C}}_{\bf{N}}}{\bf{ \times }}{{\bf{m}}_{\bf{N}}}{\bf{ \times \Delta }}{{\bf{T}}_{\bf{N}}}\). (Equation 1)

Assume the specific heat of water and coffee = 4.186\(\frac{J}{{{g^0}C}}\).

Let the final temperature be\({x^0}C\).

The temperature change for coffee = \({T_{final}} - {T_{initial}} = {x^0}C - {95^0}C\).

The density of water and coffee = 1g/mL

The volume of the coffee = 180mL

The mass of the coffee = density\( \times \)volume = 180g.

By putting the values above in equation 1, we get:

\({Q_{coffee}} = 180 \times 4.186 \times (x - 95)J\).

The temperature change for silver =\({T_{final}} - {T_{initial}} = {x^0}C - {25^0}C\).

The mass of the silver spoon = 45g

C\(_{silver}\)= 0.24\(\frac{J}{{{g^0}C}}\)(given)

\({Q_{silver}} = 45 \times 0.24 \times (x - 25)J\)

\({Q_{coffee}} + {Q_{siver}} = 0\)

Therefore,

\(180 \times 4.186 \times (x - 95)J + 45 \times 0.24 \times (x - 25)J = 0\)

\(753.48 \times (x - 95) = - 10.8 \times (x - 25)\)

\(753.48x - 71580.6 = - 10.8x + 270\)

\(753.48x + 10.8x = 71580.6 + 270\)

\(764.28x = 71850.6\)

\(x = \frac{{71850.6}}{{764.28}} = 94.01\)

We assumed that the final temperature is x.

So, the temperature of the coffee cup would be reduced by \({95^0}C - {94.01^0}C = {0.99^0}C\).