CHEM EXPERT NEEDED*** A volume of 150.0 mL of H2O is initially at 24.00 °C. A chilled steel rod at 2.50 °C is placed in the water and the final temperature of the system is 21.20 °C.

Specific heat of water = 4.184 J/(g⋅∘C) and the specific heat of steel = 0.452 J/(g⋅∘C)

Write the equation and calculate the mass of the rod.

1. m₁C₁ΔT₁ + m₂C₂ΔT₂ = 0

2. 207 g

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of one object, the same amount of heat must flow into another object.

Since there is no change in total energy,

Heat₁ + heat₂ + heat₃+ … = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂ + q₃ + … = 0

The formula for the heat absorbed or released by an object is

 q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

Heat from steel rod + heat absorbed by water = 0

               q₁               +                  q₂                    = 0

           m₁C₁ΔT₁           +            m₂C₂ΔT₂             = 0

2. Data:

m₁ = ?;               C₁ = 0.452 J°C⁻¹g⁻¹;T_f = 21.20 °C; T_i =   2.50 °C

V₂ = 150.0 mL; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 21.20 °C; T_i = 24.00 °C

3. Calculations

(a) m₂

The density of water is 0.9982 g/mL, so  

m₂ = 150.0 mL × 0.9982 g/mL = 149.7 g

(b) Temperature changes  

(c) q₁

q₁ = m₁C₁ΔT₁ = m₁ × 0.452 J°C⁻¹g⁻¹ × 18.70 °C = 8.452m₁ J·g⁻¹

(d) q₂

q₂ = m₂C₂ΔT₂ = 149.7 g × 4.184 J°C⁻¹g⁻¹ × (-2.80 °C) = -1754 J

(e) m₁

3 1/4 %

explanation:

14 days is the half-life of phosphorus-32