A piece of iron (Cp = 0.450 J/(gi°C)) that has a mass of 21.5 g and an initial temperature of 100.0°C is submerged in X g of water (Cp = 4.184 J/(gi°C)) at 20.0°C. The temperature of the system (iron and water) changes to 21.4°C. What is the value of X?

heat lost by Fe + heat gained by water = 0

[mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for mass H2O

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To solve this problem, we can use the equation:

q = m * Cp * ΔT

where:
q is the heat gained or lost by the system,
m is the mass of the substance,
Cp is the specific heat capacity of the substance, and
ΔT is the change in temperature.

First, let's calculate the heat gained or lost by the iron.

q_iron = m_iron * Cp_iron * ΔT_iron

Since the iron is submerged in water, the heat gained by the iron will be equal to the heat lost by the water.

q_iron = -q_water

Now, let's calculate the heat gained or lost by the water.

q_water = m_water * Cp_water * ΔT_water

However, we are given the initial and final temperatures of the system, so we need to calculate the change in temperature for both the iron and the water.

ΔT_iron = final temperature of the system - initial temperature of the iron
ΔT_water = final temperature of the system - initial temperature of the water

Let's substitute the known values into the equations and solve for the missing variables.

Given:
m_iron = 21.5 g
Cp_iron = 0.450 J/(g·°C)
initial temperature of the iron = 100.0°C
final temperature of the system = 21.4°C
Cp_water = 4.184 J/(g·°C)
initial temperature of the water = 20.0°C

Step 1: Calculate ΔT_iron
ΔT_iron = 21.4°C - 100.0°C
ΔT_iron = -78.6°C

Step 2: Calculate q_iron
q_iron = m_iron * Cp_iron * ΔT_iron
q_iron = 21.5 g * 0.450 J/(g·°C) * (-78.6°C)
q_iron = -737.67 J

Step 3: Calculate q_water
q_water = -q_iron
q_water = 737.67 J

Step 4: Calculate ΔT_water
ΔT_water = 21.4°C - 20.0°C
ΔT_water = 1.4°C

Step 5: Calculate m_water (mass of water)
q_water = m_water * Cp_water * ΔT_water
737.67 J = m_water * 4.184 J/(g·°C) * 1.4°C

Solving for m_water:
m_water = 737.67 J / (4.184 J/(g·°C) * 1.4°C)

m_water ≈ 131.06 g

Therefore, the value of X, which represents the mass of water, is approximately 131.06 grams.