Calculate the total amount of energy required in calories to convert 50.0 g of ice at 0.00 degrees Celsius to steam at 100. degrees Celsius. Specific heat capacity of water is 1.00 cal/g OC
Hfusion = 80 cal/g OC and Hvap = 540 cal/g OC
Write the complete equation you will use. 1 point
Substitute the values in the equation in step 1. 1 point
Report the math answer with 3 sig figs and the correct unit. 1 point

HFusion*mass + Spec.Heat*mass*ΔT + HVap*mass

80cal/g*50.0g + 1.00cal/g°C*50.0g*(100°C-0°C) + 540cal/g*50g

3.60x10⁵cal

Explanation:

Using the HFusion we can find the heat needed to convert the ice to liquid water.

With specific heat capacity we can find the heat needed to increase the temperature of water from 0 to 100°C.

With HVap we can find the heat to convert the liquid water into steam.

The equations are:

Computing the values:

36000cal =

dynamic equilibrium is the answer.