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

- lead + phosphoric acid = hydrogen + lead(ii) phosphate