achieve week 4 number 11

[Isabella Nassir 2B]

A hot lump of 37.4 g of copper at an initial temperature of 54.0 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/(g·°C)? Assume no heat is lost to surroundings.

I keep getting the answer wrong. Can someone walk through the steps please?

[Barbara Soliman 1G]

A hot lump of 37.4 g of copper at an initial temperature of 54.0 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/(g·°C)? Assume no heat is lost to surroundings.

I keep getting the answer wrong. Can someone walk through the steps please?

We need to set their q values equal to each other. Since water is absorbing heat, that q value will be positive and the q value for iron will be negative.

q(water)= -q(copper)
mc(delta T)= -(mc(delta T)
(50g)(4.184J/g⋅°C)(Tfinal - 25°C) = -(37.4)(0.385 J/ (g⋅°C)(Tfinal - 54 g°C)

Solve algebraically for Tfinal and you should get your answer!

[Andrew Nguyen 1E]

Hi,

For this one, you must consider the copper mass the system and the water the surroundings such that the heat lost from the copper will enter the water: q(copper) = - q(water).
You can the use q=mC T to get:
=

and solve for Tf!

[William Huang 1K]

You need to set the -q(metal) = q(water) because the water is absorbing heat and them solve for the Tfinal.