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A bowl containing 70 grams of water, is heated from 10 °C to 90 °C. The specific heat of

water is 4.184 J/gºC. How much heat energy is required to heat the water?

1 Answer

6 votes

Answer:

23430.4 J.

Step-by-step explanation:

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

Mass (M) = 70 g

Initial temperature (T₁) = 10 °C

Final temperature (T₂) = 90 °C

Specific heat capacity (C) = 4.184 J/gºC

Heat (Q) required =?

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

Initial temperature (T₁) = 10 °C

Final temperature (T₂) = 90 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 90 – 10

ΔT = 80 °C

Finally, we shall determine the heat energy required to heat up the water. This can be obtained as follow:

Mass (M) = 70 g

Change in temperature (ΔT) = 80 °C

Specific heat capacity (C) = 4.184 J/gºC

Heat (Q) required =?

Q = MCΔT

Q = 70 × 4.184 × 80

Q = 23430.4 J

Therefore, 23430.4 J of heat energy is required to heat up the water.

User Preeya
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