How many joules of heat are needed to heat 25.0 g of ice at -5.0 degrees celcius to a final temperature of 40.0 degrees celcius? The specific heat of ice is 2.1 j. Our teacher mention to consider the state of water present at 40.0 degrees celcius which is confusing me.

Any help is appreciated. Thanks.

You go through each state. At -5 C it is solid and goes to zero as ice.

q1 = heat absorbed in moving ice from -5C to zero.
q1 = mass ice x specific heat ice x (Tfijal-Tinitial)

q2 = heat absorbed in melting ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion.

q3 = heat absorbed in moving liquid water from zero C to 40 C.
q3 = mass water x specific heat water x (Tfinal-Tintial)
Total Q = q1 + q2 + q3.

You see this is just two things you need to worry about.
When moving WITHIN a phase you use
q = mass x specific heat x (Tfinal-Tintial)
When changing phase at a point (a certain T), one uses q = mass x delta H where delta H is Heat fusion for melting or heat vaporization for boiling. So one can go from ice at -50 C to steam at 150C and it sounds so confusing BUT it needs just those two formulas to do it as long as you remember what phase you're in and where the phase changes are, then add the individual q values together to obtain final Q.

To calculate the amount of heat energy required to heat a substance, you can use the formula:

q = m * c * ΔT

Where:
q is the amount of heat energy
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

In this case, let's break down the problem into two steps:

Step 1: Heating the ice from -5.0°C to 0.0°C
Step 2: Heating the water from 0.0°C to 40.0°C

Step 1: Calculate the heat required to change the ice into water at 0.0°C.
To do this, we need to consider the heat of fusion, which is the amount of heat energy needed to change a substance from a solid to a liquid state at its melting point.

The heat of fusion for water is 334 J/g.
The mass of the ice is given as 25.0 g.

So, to calculate the heat required to melt the ice, we use the formula:

q1 = m * ΔHf

Where:
q1 is the amount of heat energy
m is the mass of the ice
ΔHf is the heat of fusion

q1 = 25.0 g * 334 J/g
q1 = 8350 J

Therefore, 8350 Joules of heat energy are required to melt the ice.

Step 2: Calculate the heat required to heat the water from 0.0°C to 40.0°C.
Since the water is in the liquid state, we can use the specific heat capacity of water.

The specific heat capacity of water is 4.18 J/g°C.
The mass of the liquid water is 25.0 g.
The change in temperature is 40.0°C - 0.0°C = 40.0°C.

So, to calculate the heat required to heat the water, we use the formula:

q2 = m * c * ΔT

q2 = 25.0 g * 4.18 J/g°C * 40.0°C
q2 = 4180 J

Therefore, the heat required to heat the water from 0.0°C to 40.0°C is 4180 Joules.

Finally, to find the total heat required, we add the heat from both steps:

Total heat = q1 + q2 = 8350 J + 4180 J = 12530 J

Therefore, you would need 12530 Joules of heat to heat 25.0 g of ice at -5.0°C to a final temperature of 40.0°C.

To determine the amount of heat required to heat the ice, we need to consider the different steps involved:

1. Heating the ice from -5.0 degrees Celsius to 0 degrees Celsius to melt it.
2. Heating the resulting water from 0 degrees Celsius to its final temperature of 40.0 degrees Celsius.

Let's calculate the amount of heat required for each step:

1. Heating the ice to 0 degrees Celsius:
The specific heat of ice is given as 2.1 J/g°C, which means it takes 2.1 joules of heat to raise the temperature of 1 gram of ice by 1 degree Celsius.

To raise the temperature of 25.0 g of ice by 5.0 degrees Celsius, we use the formula:
Q = m * c * ΔT
where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Q1 = 25.0 g * 2.1 J/g°C * 5.0 °C
Q1 = 262.5 joules

Therefore, 262.5 joules of heat are needed to heat the ice from -5.0 degrees Celsius to 0 degrees Celsius.

2. Melting the ice at 0 degrees Celsius:
The heat of fusion for water is the amount of heat required to convert 1 gram of ice at 0 degrees Celsius to water at 0 degrees Celsius. It is given as 334 J/g.

To determine the amount of heat required to melt the ice, we use the formula:
Q2 = m * Hf
where Q2 is the heat energy, m is the mass, and Hf is the heat of fusion.

Q2 = 25.0 g * 334 J/g
Q2 = 8,350 joules

Therefore, 8,350 joules of heat are needed to melt the ice.

3. Heating the resulting water to 40.0 degrees Celsius:
Now that we have water at 0 degrees Celsius, we need to heat it to the final temperature of 40.0 degrees Celsius.

Using the specific heat of water, which is generally 4.18 J/g°C, we can calculate the amount of heat required:
Q3 = m * c * ΔT
Q3 = 25.0 g * 4.18 J/g°C * 40.0 °C
Q3 = 4,180 joules

Therefore, 4,180 joules of heat are needed to heat the water from 0 degrees Celsius to 40 degrees Celsius.

To find the total amount of heat required, we need to add together the heat required for each step:
Total heat required = Q1 + Q2 + Q3
Total heat required = 262.5 J + 8,350 J + 4,180 J
Total heat required = 12,792.5 joules

Therefore, approximately 12,792.5 joules of heat are needed to heat 25.0 g of ice at -5.0 degrees Celsius to a final temperature of 40.0 degrees Celsius, considering the state of water at 40.0 degrees Celsius.