A 18.6-g piece of gold (s = 0.129 J/(g·°C)),initially at 285.4°C,is added to 147.2 g of a liquid,initially at 27.9°C,in an insulated container.The final temperature of the metal-liquid mixture at equilibrium is 28.9°C.What is the identity of the liquid? Neglect the heat capacity of the container.

Question

Quizplus · Accepted Answer

First, we need to use the formula for calculating the heat released or absorbed during a temperature change:
q = m * s * ΔT
where q is the heat released or absorbed, m is the mass of the substance, s is its specific heat capacity, and ΔT is the temperature change.

For the gold piece, the heat released (since it is cooling down) is:
q1 = 18.6 g * 0.129 J/g°C * (285.4°C - 28.9°C) = -108356 J

For the liquid, the heat absorbed (since it is warming up) is:
q2 = 147.2 g * s * (28.9°C - 27.9°C)

At equilibrium, the total heat released must be equal to the total heat absorbed:
q1 = -q2
-108356 J = -147.2 g * s * (28.9°C - 27.9°C)
s = 4.18 J/(g·°C)

This value of specific heat capacity matches with that of water. Therefore, the liquid in the container is most likely water.

A 186-G Piece of Gold (S = 0