How much heat was transferred from 50.0 g of water if the temperature of the water went from 30.0 ° C to 55.0 °? The specific heat capacity of the water is 4.18 J/(g•K)

Taking into account the definition of calorimetry,   the heat transfered is 5225 J.

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

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

The expression for sensible heat in an isobaric process, that is, at constant pressure, is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

Replacing:

Q= 4.18 × 50 g× 25 K

Solving:

Q=5225 J

Finally, the heat transfered is 5225 J.

Learn more:

Heat transfer = Q = 62341.6 J

Explanation:

Given data:

Heat transfer = ?

Mass of water = 50.0 g

Initial temperature = 30.0°C

Final temperature = 55.0°C

Specific heat capacity of water = 4.184 J/g.K

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 55.0°C -  30.0°C

ΔT = 25°C (25+273= 298 K)

Q = 50.0 g × 4.184 J/g.K ×298 K

Q = 62341.6 J

distance, luminosity, brightness, radius, chemical composition, and temperature