Question

If 1.16 L of water is initially at 24.2 ∘C, what will its temperature be after absorption of 9.4×10−2 kWh of heat?

Answer 1

Answer:

The temperature will be 93.92 °C

Explanation:

To explain this we will use following equation: also  Q = ∆U + W known as the NON-FLOW ENERGY EQUATION (N.F.E.E.)

With Q = heat added to the system

with ∆U  = change in internal energy

⇒∆U = ( m )( Cv )( T2 - T1 )

With W = work done by the system

⇒For this situation W = 0 because there isn't work done

So we get: ∆U = ( m )( Cv )( T2 - T1 ) = Q

To find the temperature, we have to isolate T in the equation:

(T2-T1) = Q / (m)(Cv)

⇒ Since we know that m = density * volume we can calculate the mass of water.

mass = 1000g/L * 1.16 L = 1160g

Cv = heat capacity ⇒ water has a  heat capacity of 4.184 J/g °C

We know the absorption of heat is 9.4x 10^-2 kWh but to know how many joule this is we should convert ( 1 joule = 3.6 x 10^6 kWh)

⇒Q = ( 0.094 kWh ) ( 3.6 x 10^6 J / kWh ) = 0.3384 x 10^6 J

For the temperature we get then: T2 -T1 = Q / (m)(Cv)

T2 - T1 =  0.3384 x 10^6 J / (( 1160g)*(4.184 J/g °C)) = 69.72 ° C

T2 = ( T2 - T1 ) + T1   ⇒ 69.72 + 24.2 = 93.92 °C