How much heat is absorbed by a 112.5 g sample of water when it is heated from 12.5 °C to 9°C? (Specific heat capacity of water is 4.184 J/g °C)

–1647.45 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 112.5 g

Initial temperature (T₁) = 12.5 °C

Final temperature (T₂) = 9°C

Specific heat capacity (C) = 4.184 J/g°C

Heat (Q) absorbed =?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial temperature (T₁) = 12.5 °C

Final temperature (T₂) = 9°C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 9 – 12.5

ΔT = –3.5 °C

Finally, we shall determine the heat absorbed. This can be obtained as follow:

Mass (M) = 112.5 g

Change in temperature (ΔT) = –3.5 °C

Specific heat capacity (C) = 4.184 J/g°C

Heat (Q) absorbed =?

Q = MCΔT

Q = 112.5 × 4.184 × –3.5

Q = –1647.45 J

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2