A 110.0-g sample of metal at 82.00°C is added to 110.0 g of H₂O(l)at 27.00°C in an insulated container.The temperature rises to 30.56°C.Neglecting the heat capacity of the container,what is the specific heat of the metal? The specific heat of H₂O(l)is 4.18 J/(g ∙ °C).

Question

Quizplus · Accepted Answer

First, we need to calculate the heat gained by the water:
qwater = mwater * Cp,water * ΔT = 110.0 g * 4.18 J/(g ∙ °C) * (30.56°C - 27.00°C) = 1544 J

Since the container is insulated, the heat lost by the metal is gained by the water:
qmetal = -qwater = -1544 J

Now, we can use the formula for heat:
q = mmetal * Cp,metal * ΔT

Solving for Cp,metal gives:
Cp,metal = q / (mmetal * ΔT) = -1544 J / (110.0 g * (-51.44°C)) = 0.289 J/(g ∙ °C)

Therefore, the specific heat of the metal is 0.289 J/(g ∙ °C), which is choice C.

A 1100-G Sample of Metal at 82