Calculate the heat, in joules, needed to warm 225 grams of water from 88.0ºC to its boiling point, 100.0ºC.

The heat needed to warm 225 grams of water from 88.0ºC to its boiling point, 100.0ºC, is 11,296.8 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

In this way, between heat and temperature there is a direct proportional relationship (Two quantities are directly proportional when there is a constant, so that when one of the quantities increases, the other also increases; and the same happens when either of the two decreases.) . The constant of proportionality dependent on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate the calorie exchanges:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature

In this case:

Replacing:

Q= 4.184 *225 g* 12 C

Solving:

Q= 11,296.8 J

The heat needed to warm 225 grams of water from 88.0ºC to its boiling point, 100.0ºC, is 11,296.8 J

where's the diagram?

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year