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

Answer

The kilojoules of heat absorbed = 206 kJ

Explanation

Given:

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.

Step-by-step solution:

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

false

because it does not have the potential to be entirely used up

confused

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