Respuesta :

The question is incomplete, the complete question is

The following equation is the balanced combustion reaction for C6H6

[tex]2C_6H_6(l)+15O_2(g) \rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ[/tex]

If 7.400 g of [tex]C_6H_6[/tex] is burned and the heat produced from the burning is added to 5691 g of water at 21°C. What is the final temperature of the water?

Answer:

34.45°C is the final temperature of the water.

Explanation:

Moles of benzene = [tex]\frac{7.400 g}{78 g/mol}=0.09487 mol[/tex]

According to reaction, 2 moles of benzene on combustion gives 6542 kJ of heat. Then 0.09487 moles of benzene will give:

[tex]\frac{1}{2}\times 6542 kJ\times 0.09487 mol=310.325 kJ[/tex]

Heat added to water = Q = 310.325 kJ = 310,325 J

(1 kJ = 1000 J)

Specific heat of water = C = [tex]4.18 J/g^oC[/tex]

Mass of the water = m = 5691 g

Initial temperature of the water =[tex]T_1[/tex] = 21°C

Final temperature of the water =[tex]T_2[/tex] =?

Change in temperature of the substance =ΔT =[tex]T_2-T_1[/tex]

[tex]Q=mc\times \Delta (T_2-T_1)[/tex]

[tex]310,325 J=5691 g\times 4.18 J/g^oC\times (T_2-21^oC)[/tex]

[tex]T_2=34.45^oC[/tex]

34.45°C is the final temperature of the water.

Answer:

The final temperature of the water is 34.0 °C

Explanation:

Step 1: Data given

Mass of C6H6 burned = 7.400 grams

Mass of water = 5691 grams

Temperature of water = 21 °C

Specific heat of water = 4.184 K/g°C

2C6H6+15O2→12CO2+6H20+6542KJ

Heat released with combustion of C6H6 = 6542KJ

Step 2: Calculate moles of C6H6

Moles C6H6 = mass C6H6 / molar mass C6H6

Moles C6H6 = 7.400 grams / 78.11 g/mol

Moles C6H6 = 0.09474

Step 3: Calculate heat transfered

Q = 0.09474 moles * 6542 kJ/2 mol = 309.89 kJ

Step 3: Calculate finam temperature

Q = m * c * ΔT

⇒ with Q = the heat transfered =309.9 kJ = 309890 J

⇒ with m = the mass of water =  5691 grams

⇒ with c = the specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature

ΔT = Q /(m*c)

ΔT = 309890 / ( 5691 * 4.184)

ΔT =13.0 °C

Step 4: Calculate the final temperature

T2 = T1 + ΔT

T2 = 13.0 + 21.0°C

T2 = 34.0 °C

The final temperature of the water is 34.0 °C