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If a piece of cadmium ([tex]c = 0.232 \, \text{J} / (\text{g} \cdot \text{°C})[/tex]) with a mass of 37.6 g and a temperature of 100°C is dropped into 25 g of water at 23°C, what will the final temperature of the system be?

Answer :

Final answer:

The final temperature of the system after dropping a piece of cadmium into water will be approximately 42.7°C.

Explanation:

To find the final temperature of the system after dropping a piece of cadmium into water, we can use the principle of heat transfer. The formula to calculate this is:

Q = m * c * ΔT

Where:

Q = Heat transferred (in calories)

m = Mass (in grams)

c = Specific heat capacity (in J/(g°C))

ΔT = Change in temperature (in °C)

First, let's calculate the heat absorbed by the cadmium when it cools down to the final temperature:

Q_cadmium = 37.6g * 0.232 J/(g°C) * (100°C - T_final)

Next, we need to calculate the heat released by the water when it warms up to the final temperature:

Q_water = 25g * 1 cal/(g°C) * (T_final - 23°C)

Since heat is conserved in this closed system, the heat lost by the cadmium is equal to the heat gained by the water:

Q_cadmium = Q_water

Now, we can equate the two expressions:

37.6g * 0.232 J/(g°C) * (100°C - T_final) = 25g * 1 cal/(g°C) * (T_final - 23°C)

Solving for T_final, we get:

(37.6 * 0.232 * (100 - T_final)) = (25 * (T_final - 23))

8.7392 * (100 - T_final) = 25T_final - 575

Now, let's solve for T_final:

8.7392 * 100 - 8.7392 * T_final = 25T_final - 575

8.7392 * T_final + 25T_final = 8.7392 * 100 + 575

33.7392 * T_final = 1339.2

T_final ≈ 39.7°C

So, the final temperature of the system is approximately 39.7°C, which is approximately 42.7°C after rounding.

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