Question

A new 125 g alloy of brass at 100°C is dropped into 76 g of water at 25 °C. The final temperature of the water and brass is 35 °C, what is the specific heat of the sample of brass? The specific heat of water = 4.184 J/g. °C​

Answer 1

The specific heat of the brass can be calculated using the formula:

Q = mcΔT

where Q is the heat transferred, m is the mass of the brass, c is the specific heat of the brass, and ΔT is the change in temperature.

First, calculate the heat transferred from the brass to the water:

Qbrass = mcΔT = (125 g)(c)(100 °C - 35 °C) = 9375c J

Next, calculate the heat transferred from the water to the brass:

Qwater = mcΔT = (76 g)(4.184 J/g. °C)(35 °C - 25 °C) = 3191.84 J

Since the heat lost by the brass is equal to the heat gained by the water:

Qbrass = Qwater

9375c J = 3191.84 J

c = 0.34 J/g. °C

Therefore, the specific heat of the brass is 0.34 J/g. °C.

Explanation: