A student must use 225 g of hot water in a lab procedure. Calculate the amount of heat in joules required to raise the temperature of 225 g of water from 20.0 °C to 100.0 °C. (Water’s specific heat capacity is 4.184 J/g°C) (round answer to 2-3 sig figs)

Q = 75.312 KJ

Explanation:

Heat is the amount of the thermal energy contained in an object measured in joules.  When heat energy is transferred to an object, the temperature of the object changes depending on the amount of heat applied and the type of object.

Given that:

mass of water (m) = 225 g, Temperature difference () = 100°C - 20°C = 80°C, Water’s specific heat capacity () = 4.184 J/g°C

The quantity of heat (Q) is given as:

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.