Question

A 92.2 g sample of gold at 100.0°C is placed in 100.0 g

water at 21.0°C; the final temperature of the gold and
water is 23.2°C. What is the specific heat of gold? ( can you show how to do it)

Answer 1

To find the specific heat of gold, we can use the equation: q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat, and ΔT is the change in temperature. Using the given information, we can calculate the specific heat of gold to be 0.129 J/g °C.

Step-by-step explanation:

To find the specific heat of gold, we can use the equation:

q = mcΔT

where q is the heat absorbed or released, m is the mass, c is the specific heat, and ΔT is the change in temperature.

In this case, we are given the mass of the gold (92.2 g), the initial temperature (100.0°C), the final temperature (23.2°C), and the specific heat of gold (0.129 J/g °C).

Using the formula, we can calculate:

q = mcΔT = (92.2 g)(0.129 J/g °C)(23.2°C - 100.0°C)

q = -1132.1334 J

The negative sign indicates that heat is lost by the gold.

Since heat lost by the gold is equal to the heat gained by the water, we can use the equation:

q = mcΔT

to find the mass of the water.

Let's assume the specific heat of water is 4.18 J/g °C.

Using the equation, we have:

(92.2 g)(0.129 J/g °C)(100.0°C - 23.2°C) = (m)(4.18 J/g °C)(23.2°C - 21.0°C)

Solving for m, the mass of the water, we find:

m = 270.298 g

Therefore, the specific heat of gold is 0.129 J/g °C.