Calculate the final temperature of a 63.7 g piece of palladium,originally at 100.0 °C,placed inside a coffee cup calorimeter containing 115.0 mL of water at 21.8 °C.The specific heat capacity of water is 4.184 J g^-1 °C^-1,the specific heat capacity of palladium is 0.246 J g^-1 °C^-1,and the density of water is 1.00 g mL^-1.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat exchanged between the palladium and the water:

q = mPd * CPd * ΔT = -(mH2O * Cp * ΔT)

where
- q is the heat exchanged (in Joules)
- mPd is the mass of palladium (in grams)
- CPd is the specific heat capacity of palladium (in J/g°C)
- mH2O is the mass of water (in grams)
- Cp is the specific heat capacity of water (in J/g°C)
- ΔT is the change in temperature (in °C)

Solving for ΔT, we get:

ΔT = - q / (mH2O * Cp) = mPd * CPd / (mH2O * Cp)

Plugging in the given values:

ΔT = (63.7 g) * (0.246 J/g°C) / [(115.0 g) * (4.184 J/g°C)] = -0.784°C

Note: the negative sign indicates that the palladium lost heat to the water, causing its temperature to decrease.

Finally, we can calculate the final temperature of the water and the palladium:

Tw = Ti + ΔT = 21.8°C - 0.784°C = 21.0°C
Tp = Ti + ΔT = 100.0°C - 0.784°C = 99.2°C

Therefore, the final temperature of the palladium is:

Answer: D) 24.3°C

Calculate the Final Temperature of a 63