Calculate the final temperature of 202 mL of water initially at 32 degrees celsius upon absorption of 19 kJ of heat

The  Final Temperature of water = 54.48°C.

Explanation:

Given,

Volume of water = 202 mL.

Initial Temperature T₁ = 32°C

Heat energy absorbed = 19 kJ = 19000 Joules

Final Temperature T₂ = ?

Final temperature can be calculated using the formula,

Q = (mass) (ΔT) (Cp)

Here  ΔT = T₂ - T₁

Cp = specific heat of water = 4.184 J/g°C.

For finding the mass, we use the formula,

Density = Mass/Volume

Since, Density of water =  1 g/mL.

Mass of water = 202 grams.

Now Substituting all values in the formula Q = (mass) (ΔT) (Cp)

19000 = (202)(T₂ - T₁)(4.184)

Solving, we get,

(T₂ - T₁) = 22.48°C

Therefore,  T₂ = 22.48°C + 32°C = 54.48°C.

Therefore, Final Temperature of water = 54.48°C.

i don,t know so just answer mine