How many joules of heat are absorbed to raise the temperature of 125 grams of water at 1 atm from 35°C to its boiling point, 100.°C.

33962.5 J

Explanation:

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

Mass (M) = 125 g

Initial temperature (T₁) = 35 °C

Initial temperature (T₂) = 100 °C

Specific heat capacity (C) = 4.18 J/gºC

Heat (Q) absorbed =?

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

Initial temperature (T₁) = 35 °C

Initial temperature (T₂) = 100 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 100 – 35

ΔT = 65 °C

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

Mass (M) = 125 g

Change in temperature (ΔT) = 65 °C

Specific heat capacity (C) = 4.18 J/gºC

Heat (Q) absorbed =?

Q = MCΔT

Q = 125 × 4.18 × 65

Q = 33962.5 J

Therefore, the heat absorbed is 33962.5 J

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