Question

A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24°C is placed in 180 mL (180 g) of coffee at 85°C and the temperature of the two becomes equal.

(a) What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water.

(b) The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer.

Assume that the coffee has the same density and specific heat as water.

Short answer

a. The final temperature of the coffee cup is 81.95℃.

b. The solution is incorrect because the temperature would be less than 88℃.

Step-by-step solution

Heat change

Under ideal circumstances, the net heat change is zero.

\({q_{subs\tan ceM}} + {q_{subs\tan ceN}}\) = 0

This relationship can be rearranged to show that the heat gained by the substance

M is equal to the heat lost by the 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

Let us assume that the substance M is coffee and the substance N is aluminium.

\({Q_M} = {C_M} \times {m_M} \times \Delta {T_M}\)and \({Q_N} = {C_N} \times {m_N} \times \Delta {T_N}\)

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

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

The temperature change for the coffee = \({T_{final}} - {T_{initial}} = {x^0}C - {85^0}C\)

The density of the water and the coffee = 1g/mL

The volume of the coffee = 180mL

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

Putting the above values in the equation we get,

\({Q_{coffee}} = 180 \times 4.186 \times (x - 85)J\)

Calculation of the final temperature

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

The mass of the silver spoon = 45g.

C\(_{alu\min um}\)= 0.88\(\frac{J}{{{g^0}C}}\) [given]

\({Q_{silver}} = 45 \times 0.88 \times (x - 24)J\)

\({Q_{coffee}} + {Q_{alu\min um}} = 0\)

Therefore,

\(180 \times 4.186 \times (x - 85)J + 45 \times 0.88 \times (x - 24)J = 0\)

\(753.48 \times (x - 85) = - 39.6 \times (x - 24)\)

\(753.48x - 64045.8 = - 39.6x + 950.4\)

\(753.48x + 39.6x = 64045.8 + 950.4\)

\(793.08x = 64996.2\)

\(x = \frac{{64996.2}}{{793.08}} = 81.95\)

We assumed in Step 2 to let the final temperature be x℃.

So, the final temperature of the cup becomes 81.95℃.

Explanation of the incorrect answer

It is clear that the temperature of the whole system would decrease due to the fact that the heat would flow from higher temperatures towards lower temperatures.

Since the higher temperature is 85℃, therefore, the final temperature would always be less than 85℃. Hence, the solution made by the student is incorrect.