How much heat is required to raise the temperature of 90.0g of water from its melting point to its boiling point?

Can someone please help me set this up, I am so confused

heat=mass*specificheatcapacity*changeinTemp

heat=90*c*(100C-0C)

look up the value of specific heat c, and you have it.

All of these are worked with that magical formula of

q = mc*delta T
q = 90 x specific heat H2O x (Tfinal-Tinitial)
specific heat H2O is 4.184 J/g*C
Tf = 100
Ti = 0

To calculate the amount of heat required to raise the temperature of water from its melting point to its boiling point, we need to consider the specific heat capacity of water and the phase changes involved.

Let's break down the process step by step:

Step 1: Determine the heat required to raise the temperature of water from its melting point (0°C) to its boiling point (100°C) using the formula:

Q1 = mass × specific heat capacity × ΔT1

Where:
Q1 is the heat required
mass is the mass of water
specific heat capacity is the specific heat capacity of water (4.18 J/g°C)
ΔT1 is the change in temperature (100°C - 0°C)

Given:
Mass of water (m) = 90.0 g
Specific heat capacity of water (c) = 4.18 J/g°C
Change in temperature (ΔT1) = 100°C - 0°C = 100°C

Substituting these values into the equation, we get:

Q1 = 90.0 g × 4.18 J/g°C × 100°C
Q1 = 37620 J

Step 2: Determine the heat required for the phase change from solid to liquid (melting) using the formula:

Q2 = mass × heat of fusion

The heat of fusion for water is 334 J/g.

Given:
Mass of water (m) = 90.0 g
Heat of fusion (ΔHfus) = 334 J/g

Substituting these values into the equation:

Q2 = 90.0 g × 334 J/g
Q2 = 30060 J

Step 3: Determine the heat required for the phase change from liquid to gas (vaporization) using the formula:

Q3 = mass × heat of vaporization

The heat of vaporization for water is 2260 J/g.

Given:
Mass of water (m) = 90.0 g
Heat of vaporization (ΔHvap) = 2260 J/g

Substituting these values into the equation:

Q3 = 90.0 g × 2260 J/g
Q3 = 203400 J

Step 4: Calculate the total heat required by summing up the heat from each step:

Total Heat = Q1 + Q2 + Q3
Total Heat = 37620 J + 30060 J + 203400 J
Total Heat = 271080 J

Therefore, it would require 271080 Joules of heat to raise the temperature of 90.0 grams of water from its melting point to its boiling point.

To calculate the heat required to raise the temperature of a substance, you can use the specific heat formula:

Q = m * c * ΔT

Where:
Q is the heat energy (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C), and
ΔT is the change in temperature (in °C).

In this case, we need to calculate the heat required to raise the temperature of 90.0g of water from its melting point to its boiling point.

First, let's determine the ΔT (change in temperature) from the melting point to the boiling point of water. The boiling point of water is 100°C, and the melting point is 0°C.

ΔT = boiling point - melting point
ΔT = 100°C - 0°C
ΔT = 100°C

Now we need to find the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.

Now we can substitute the values into the formula:

Q = m * c * ΔT
Q = 90.0g * 4.18 J/g°C * 100°C

Let's calculate the result:

Q = 90.0g * 4.18 J/g°C * 100°C
Q = 37,620 J

Therefore, approximately 37,620 joules of heat energy are required to raise the temperature of 90.0 grams of water from its melting point to its boiling point.