Answer:
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 the reactor (negative), the same amount of heat must flow into the water (positive).
Since there is no change in total energy,
heat₁ + heat₂ = 0
The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as
q₁ + q₂ = 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 released by reactor + heat absorbed by water = 0
q₁ + q₂ = 0
q₁ + m₂C₂ΔT₂ = 0
2. Data:
q₁ = -23 746 kJ
m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 95 °C; T_i = 10 °C
3. Calculations
(a) Convert kilojoules to joules
(b) ΔT
ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C
(c) m₂
