A 1.00-kg piece of iron at 655°C is dropped into a calorimeter of negligible heat capacity containing 50.1 g of water at 30°C. What is the final temperature of the system? The specific heat of iron is 448 J/(kg K). the specific heat of steam is 2010 J/(kg K). and the latent heat of vaporization of water is 22.6 × 10⁵ J/kg.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat released by the iron:

q = m x c x ΔT
q = 1 kg x 448 J/(kg K) x (655°C - T)
q = 448(655-T) J

The heat gained by the water can be calculated using:

q = m x c x ΔT
q = 50.1 g x 4.18 J/(g K) x (T - 30°C)
q = 2098.86(T-30) J

Since there is no heat lost or gained by the calorimeter, we can set the two heat equations equal to each other:

448(655-T) = 2098.86(T-30)

Solving for T, we get:

T = 319.8°C

Therefore, the final temperature of the system is approximately 320°C, which means the answer is C.

A 100-Kg Piece of Iron at 655°C Is Dropped into a a Calorimeter