Question

A 33.41 g piece of copper at 95.6 Celsius was placed in a sample of water at 21.65 Celsius. The temperature of the water increased to 22.87 Celsius. To what mass of water was the cooper added? The specific heat of copper is 0.385 J/g C

Answer 1

43.63 g of water.

Step-by-step explanation:

We can use the equation for heat transfer to solve for the mass of water:

q = cmΔT

where

q = heat transfer

c = specific heat

m = mass

ΔT = change in temperature

The heat transfer from the copper to the water can be expressed as:

q = cmΔT

q = (33.41 g)(0.385 J/g°C)(95.6°C - 21.65°C)

And the heat transfer caused a change in temperature in the water, which can be expressed as:

q = mLΔT

q = (m)(4.18 J/g°C)(ΔT)

Equating the two expressions for heat transfer, we can solve for the mass of water:

(33.41 g)(0.385 J/g°C)(95.6°C - 21.65°C) = (m)(4.18 J/g°C)(22.87°C - 21.65°C)

m = (33.41 g)(0.385 J/g°C)(95.6°C - 21.65°C) / (4.18 J/g°C)(22.87°C - 21.65°C)

m = 43.63 g

So the copper was added to 43.63 g of water.