How many calories are absorbed when 150g of ice at O is changed to liquid water at 45C?

q1 = heat to melt ice at zero C to liquid at zero C.

q1 = mass ice x heat fusion.

q2 = heat to warm water at zero C to 45 C.
q2 = mass water x specific heat H2O x (Tfinal-Tinitial).

Total = q1 + q2.

To determine the number of calories absorbed when 150g of ice at 0°C is changed to liquid water at 45°C, you need to consider the energy required for two processes: raising the temperature of the ice to its melting point, and then melting the ice into liquid water.

First, let's calculate the calories required to raise the temperature of the ice from 0°C to its melting point, which is also 0°C. The specific heat capacity of ice is 0.5 calories per gram per degree Celsius (cal/g°C).

The equation to calculate the calories required to raise the temperature of the ice is:

q = m * c * ΔT

where:
q is the heat energy (in calories),
m is the mass of the ice (in grams),
c is the specific heat capacity of ice (0.5 cal/g°C),
ΔT is the change in temperature (which is 0°C - 0°C, so it is 0).

Plugging in the values, the equation becomes:

q = 150g * 0.5 cal/g°C * 0°C
q = 0 calories

Since there is no change in temperature, no calories are absorbed in raising the temperature of the ice.

Next, let's calculate the calories required to melt the ice into liquid water at 0°C. The heat of fusion, also known as the latent heat of fusion, for ice is 79.7 calories per gram (cal/g).

The equation to calculate the calories required to melt the ice is:

q = m * ΔH

where:
q is the heat energy (in calories),
m is the mass of the ice (in grams),
ΔH is the heat of fusion (79.7 cal/g).

Plugging in the values, the equation becomes:

q = 150g * 79.7 cal/g
q = 11,955 calories

Therefore, 11,955 calories are absorbed when 150g of ice at 0°C is changed to liquid water at 0°C.