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

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

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