Final answer:
The heat required to melt the ice can be calculated using the formula Q = m * AH, where m is the mass of the ice .The total heat required to warm 50.0 g of ice from -10.0°C to 120.0°C is 152,660 J.
Step-by-step explanation:
To calculate the total heat required, we need to consider the different steps involved:
- Heat to warm the ice from -10.0°C to 0.00°C:
- Using the formula Q = m * C * AT, where m is the mass of the ice (50.0 g), C is the specific heat capacity of ice (2.06 J/g°C), and AT is the change in temperature (0.00°C - (-10.0°C) = 10.0°C), we can calculate the heat required as Q1 = 50.0 g * 2.06 J/g°C * 10.0°C = 1030.0 J.
Melting the ice:
- The heat required to melt the ice can be calculated using the formula Q = m * AH, where m is the mass of the ice (50.0 g) and AH is the heat of fusion for ice (334 J/g). Therefore, Q2 = 50.0 g * 334 J/g = 16700 J.
Heating the water from 0.00°C to 100.0°C:
- Using the formula Q = m * C * AT, where m is the mass of the water (50.0 g), C is the specific heat capacity of water (4.18 J/g°C), and AT is the change in temperature (100.0°C - 0.00°C = 100.0°C), we can calculate the heat required as Q3 = 50.0 g * 4.18 J/g°C * 100.0°C = 20900 J.
Boiling the water:
- The heat required to boil the water can be calculated using the formula Q = m * AH, where m is the mass of the water (50.0 g) and AH is the heat of vaporization for water (2260 J/g). Therefore, Q4 = 50.0 g * 2260 J/g = 113000 J.
Heating the steam from 100.0°C to 120.0°C:
- Using the formula Q = m * C * AT, where m is the mass of the steam (50.0 g), C is the specific heat capacity of steam (2.03 J/g°C), and AT is the change in temperature (120.0°C - 100.0°C = 20.0°C), we can calculate the heat required as Q5 = 50.0 g * 2.03 J/g°C * 20.0°C = 2030.0 J.
The total heat required is the sum of all the individual heats: Q_total = Q1 + Q2 + Q3 + Q4 + Q5
= 1030.0 J + 16700 J + 20900 J + 113000 J + 2030.0 J
= 152660 J.