Calculate the energy needed to transform a 20 gram block of ice at -15 degrees into water vapor at 150 degrees.

To calculate the total energy needed, we need to consider the three phases of the transformation separately:

1. From ice at -15°C to ice at 0°C (raising the temperature):
The specific heat capacity of ice is c_{ice} = 2.09 \, \frac{J}{g \cdot °C} .
The energy required is given by the formula: Q = mc\Delta T , where
m = mass of the ice (20g),
c = specific heat capacity of ice (2.09 \frac{J}{g \cdot °C} ) and
\Delta T = change in temperature = 0°C - (-15°C) = 15°C.
Substitute these values into the formula:
Q_{1} = 20g \times 2.09 \frac{J}{g \cdot °C} \times 15°C = 627J .

2. From ice at 0°C to water at 0°C (melting):
The specific heat capacity of ice is c_{fusion} = 334 \, \frac{J}{g} .
The energy required is: Q_{2} = m \times c_{fusion} = 20g \times 334 \frac{J}{g} = 6680J .

3. From water at 0°C to water vapor at 150°C (raising the temperature):
The specific heat capacity of water is c_{water} = 4.18 \, \frac{J}{g \cdot °C} .
The energy required is: Q_{3} = m \times c_{water} \times \Delta T ,
where \Delta T = 150°C - 0°C = 150°C.
Substitute these values into the formula:
Q_{3} = 20g \times 4.18 \frac{J}{g \cdot °C} \times 150°C = 12540J .

Add the energies required for each step to find the total energy:
Total \, Energy = Q_{1} + Q_{2} + Q_{3} = 627J + 6680J + 12540J = 19847J .

\textbf{Answer: The total energy needed to transform the 20 gram block of ice at -15°C into water vapor at 150°C is 19847J.}