If 621000 joules of energy are added to 3.30 liters of water at 286 kelvin what will the final temperature of the water be? temperature in kelvin

Explanation:To convert from cal/(g*C) to J/(kg*K), we just need to find a conversion factor for specific heat. There is really no mathematical way to do this other than to look in a physics or chemistry book and find a conversion factor. After doing this, you will see that 1 cal/(g*C) is equal to 4,186 J/(kg*K).

To find the specific heat of a material, first look at the units. There is energy per unit mass per unit temperature. So if we are given an amount of energy appllied to an object, its mass and how much the temperature of the object rises, we can calculate its specifc heat by dividing the energy by both the mass and the temperature, but don't forget to keep the units as they are:

Specific heat of the metal = (95 cal)/(10 K * 700g) = 0.014 cal/(g*K)

To find how much energy it requires to melt 250 grams of ice, we will need what is called the Latent Heat of Melting for ice. This is the amount of heat required to change unit mass of a solid into unit mass of a liquid at a constant temperature. Again, using a reference, the latent heat of melting for ice is found to be 334 kJ/kg. So the energy required to melt one kg of ice is 334 kJ. The amount of energy required to melt 0.250 kg of ice is then:

334 * 0.250 = 83.5 kJ

i think it is true since you can test the answer and can use observations to complete it.

6.4 cm (max) - 25 cm (min)