How many kilojoules of heat are absorbed when 950 mL of water is heated from 27 degrees Celsius to 79 degrees Celsius? The specific heat of water is 4.18 J/g-C. Answer choices: 104 kJ 12 kJ 120 kJ 206 kJ
The kilojoules of heat absorbed = 206 kJ
Volume of water = 950 mL
Initial temperature, T₁ = 27 ⁰C
Final temperature, T₂ = 79 ⁰C
Specific heat of water, c = 4.18 J/g ⁰C
What to find:
The kilojoules of heat absorbed.
Since density of water = 1 g/mL
Then the mass, m of water can be calculated first.
Density = Mass/Volume
⇒ Mass, m = Density x Volume = 1 g/mL x 950 mL = 950 g
Also, ΔT = T₂ - T₁ = 79 ⁰C - 27 ⁰C = 52 ⁰C
Therefore the kilojoules of heat absorbed, Q can be calculated using the formula given below
Therefore, the kilojoules of heat absorbed is 206 kJ
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the dissolution of polymer can be thermodynamically described in 2 steps.