High School

How much energy is required to raise the temperature of 10.9 grams of water from 27.9 degrees Celsius to 38.2 degrees Celsius?

The specific heat capacity of liquid water is 4.184 J/g°C.

Answer :

Final answer:

To find the energy required to heat water, we use the formula q=m*c*ΔT. Inserting the known values, we find that 467.9 Joules of energy is required to raise the temperature of 10.9 grams of water from 27.9 degrees Celsius to 38.2 degrees Celsius.

Explanation:

The question involves the calculation of energy required to heat up water, which is a common question in thermal physics. This can be calculated using a formula q=m*c*ΔT, where 'q' is the energy required, 'm' is the mass, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature. Here, the mass 'm' is 10.9 grams, the specific heat 'c' of water is 4.184 J/g°C, and the change in temperature 'ΔT' is (38.2-27.9)°C.

Therefore, q=10.9g*4.184J/g°C*(38.2°C-27.9°C)=467.9J. So, 467.9 Joules of energy is required to heat 10.9 grams of water from 27.9 degrees Celsius to 38.2 degrees Celsius.

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