Question

How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver spoon (specific heat 0.24 J/g °C) at 25 °C is placed in the coffee, and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water.

Short answer

The temperature of the coffee cup is reduced by 0.99℃.

Step-by-step solution

Heat change

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.

Specific heat of substances M and N

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:

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

Calculation of coffee cup

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 \({95^0}C - {94.01^0}C = {0.99^0}C\).