Calculate the amount of heat required to take 3.00kg of ice at -10 degrees Celsius to 115 degrees celsius.
break this up into steps
ice from -10 to 0
melting the ice at 0
heating the water from 0 to 100
vaporizing the water at 100
heating the steam from 100 to 115
figure all the heats, then add.
To calculate the amount of heat required to take ice at -10 degrees Celsius to 115 degrees Celsius, we need to consider the different phases and temperature ranges involved. Here's how you can approach it:
1. Determine the heat required to raise the temperature of the ice from -10 degrees Celsius to its melting point at 0 degrees Celsius.
The specific heat capacity of ice (c) is 2.09 J/g°C. However, since the given mass is in kilograms, we need to convert it:
Mass of ice (m) = 3.00 kg = 3000 g
The heat required Q1 can be calculated using the formula:
Q1 = mcΔT1
Where,
m = mass of ice
c = specific heat capacity of ice
ΔT1 = change in temperature from -10°C to 0°C
ΔT1 = 0°C - (-10°C) = 10°C
Q1 = (3000 g) x (2.09 J/g°C) x (10°C)
2. Determine the heat required to melt the ice at its melting point (0 degrees Celsius).
The heat of fusion for ice (L) is 334 J/g. Hence, the heat required Q2 can be calculated using the formula:
Q2 = mL
Q2 = (3000 g) x (334 J/g)
3. Determine the heat required to raise the temperature of the water from 0 degrees Celsius to 115 degrees Celsius.
The specific heat capacity of water (c) is 4.18 J/g°C. Now we have water instead of ice, so we use this value.
The heat required Q3 can be calculated using the formula:
Q3 = mcΔT3
Where,
m = mass of water (which is equal to the original mass of ice)
c = specific heat capacity of water
ΔT3 = change in temperature from 0°C to 115°C
ΔT3 = 115°C - 0°C = 115°C
Q3 = (3000 g) x (4.18 J/g°C) x (115°C)
To get the total heat required, the individual quantities Q1, Q2, and Q3 need to be added together:
Total Heat Required = Q1 + Q2 + Q3
Please calculate the values from these equations to find the total amount of heat required.