How much mechanical energy is required to increase the temperature of 300g of water from 20C to 95C?

Answer 1

You are going to need a value for heat capacity of water, which tells us, how much energy is needed to heat 1 g of water by 1 degree of Kelvin scale (which can be considered the same as Celsius in this case, since we are only looking at temperature difference).

For liquid water at this temperature range, the value is:

cp = 4.1813 J/(g*K)

The amount of mechanical energy needed to heat the water is then:

E = ΔT * m * cp , where:

E - amount of mechanical energy,

m - mass of heated substance (in grams),

ΔT - temperature difference [K]; in this case: ΔT = 95 K - 20 K = 75 K,

cp - heat capacity [J/(g*K)].

After plugging in the numbers, we get:

E = 75 K * 300 g * 4.1813 J/g*K = 94 079.25 J,

which is about the energy released on impact from 1 ton weight falling 10 meters - and that's assuming there is no dissipation.

Answer 2

The energy required can be calculated using the specific heat capacity of water (4.18 J/g°C). The formula is: Energy = mass * specific heat capacity * temperature change. Plugging in the values, we get: Energy = 300g * 4.18 J/g°C * (95°C - 20°C). Calculating this gives an answer of approximately 95,100 Joules.