what is the specific heat of a 25.0 g substance that absorbs 493.4 j and raises the temperature from 12.0°c to 34°c?

Specific heat of substance  = 0.897 J/g.°C

Explanation:

Given data:

Specific heat of substance = ?

Mass of substance = 25.0 g

Heat absorbed = 493.4 J

Initial temperature = 12.0 °C

Final temperature = 34°C

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 = 34°C -12.0°C

ΔT = 22°C

493.4 J  = 25 g ×c× 22°C

493.4 J  = 550 g.°C ×c

c = 493.4 J  / 550 g.°C

c = 0.897 J/g.°C

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation: