How many grams of Fe can be heated from 40.0oC to 60.0oC when 20.0 g of Cu cools from 100.0oC to 25.0oC?

Specific heat of Fe = 0.444 J/(g oC); specific heat of Cu = 0.382 J/(g oC)

heat lost by Cu + heat gained by Fe = 0

[mass Cu x specific heat Cu x (Tf-Ti)] + [mass Fe x specific heat Fe x (Tf-Ti) = 0
Substitute and solve for mass Fe
Post your work if you get stuck.

To solve this problem, we need to use the formula for heat transfer:

q = m × c × ΔT

where:
q is the heat transfer
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

We have the following information:
For Fe:
Initial temperature (T1) = 40.0oC
Final temperature (T2) = 60.0oC
Specific heat (c) = 0.444 J/(g oC)

For Cu:
Initial temperature (T1) = 100.0oC
Final temperature (T2) = 25.0oC
Mass (m) = 20.0 g
Specific heat (c) = 0.382 J/(g oC)

First, let's calculate the heat transfer for Cu:

q1 = m × c × ΔT
ΔT1 = T2 - T1
q1 = 20.0 g × 0.382 J/(g oC) × (25.0oC - 100.0oC)

Next, let's calculate the heat transfer for Fe:

q2 = m × c × ΔT
ΔT2 = T2 - T1
We need to find the mass of Fe that can be heated, so we'll let the mass of Fe be represented by m2:
q2 = m2 × 0.444 J/(g oC) × (60.0oC - 40.0oC)

Since the heat transfer for the system is conserved (energy is neither created nor destroyed), we can assume that the heat lost by Cu is equal to the heat gained by Fe:

q1 = q2

Now, we can solve for m2:

20.0 g × 0.382 J/(g oC) × (25.0oC - 100.0oC) = m2 × 0.444 J/(g oC) × (60.0oC - 40.0oC)

By solving this equation, we can find the mass of Fe that can be heated from 40.0oC to 60.0oC.