what amount of energy in calories would it take to bring 20 grams of solid ice at negative twelve degrees celsius to liquid water at 100 degrees celsius?
Q = ΔU1 + ΔU2 + ΔU3 =
=c1•m •ΔT1 +λ•m + c2•m •ΔT2 =
=20•10^-3•(2.060•10^3• 12 + 335•10^3 + 4.183•10^3•100) =
=15560 J
(c1, c2 –specific heats of ice and water, λ –heat of fusion of ice)
To calculate the amount of energy in calories required to bring solid ice at -12 degrees Celsius to liquid water at 100 degrees Celsius, we need to consider two processes:
1. Heating the ice from -12 degrees Celsius to 0 degrees Celsius
2. Melting the ice at 0 degrees Celsius to liquid water at 100 degrees Celsius
Here's how you can calculate the energy required for each step:
Step 1: Heating the ice from -12 degrees Celsius to 0 degrees Celsius:
The specific heat capacity of ice is 0.5 calories per gram per degree Celsius. We're given that the mass of the ice is 20 grams. So, the energy required to heat the ice from -12 degrees Celsius to 0 degrees Celsius can be calculated using the following equation:
Energy = mass × specific heat capacity × change in temperature
Energy = 20 g × 0.5 calories/g°C × (0°C - (-12°C))
Energy = 20 g × 0.5 calories/g°C × 12°C
Step 2: Melting the ice at 0 degrees Celsius to liquid water at 100 degrees Celsius:
The heat of fusion (also known as the latent heat) of ice is 79.7 calories per gram. We're still dealing with 20 grams of ice converting it to liquid water at 100 degrees Celsius. Therefore, the energy required to melt the ice can be calculated using the equation:
Energy = mass × heat of fusion
Energy = 20 g × 79.7 calories/g
Now, to find the total energy required, add the energies calculated in each step:
Total Energy = Heating Energy + Melting Energy
Total Energy = Energy required to heat ice + Energy required to melt ice
Once you substitute the values you calculated, you will find the amount of energy in calories required to bring 20 grams of solid ice at -12 degrees Celsius to liquid water at 100 degrees Celsius.