Final answer:
The total amount of heat absorbed by 0.1 g of ice at -20.0°C to become water at 0.0°C is 0.119635 J.
Step-by-step explanation:
In order to calculate the heat absorbed, we need to consider the different stages of the process: 1) raising the temperature of the ice from -20.0°C to 0.0°C, 2) melting the ice to water at 0.0°C.
The first stage requires calculating the heat using the equation Q = mcΔT, where m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature. The second stage requires calculating the heat using the equation Q = mlf, where m is the mass of the ice, and lf is the latent heat of fusion for ice.
Let's calculate these values:
- First stage:
m = 0.1 g = 0.0001 kg
c = 2.06 cal/g °C = 4.314 J/g °C
ΔT = 0.0°C - (-20.0°C) = 20.0°C
Q = (0.0001 kg)(4.314 J/g °C)(20.0°C) = 0.08628 J - Second stage:
m = 0.1 g = 0.0001 kg
lf = 79.8 cal/g = 333.55 J/g
Q = (0.0001 kg)(333.55 J/g) = 0.033355 J
Therefore, the total amount of heat absorbed by 0.1 g of ice is 0.08628 J + 0.033355 J = 0.119635 J.