Calculate the heat that must be supplied to a 544 g stainless steel vessel containing 333 g of water to raise its temperature from 16.4◦C to the boiling point of water 100◦C

**Please help me find the delta T of steel too**

For the water:

(mass H2O x specific heat H2O x (Tfinal-Tinitial)
You have mass, sp.h., Tf and Ti. Substitute and solve for q. Use 4.184 J/g*C for sp.h. H2O

For the vessel:
There are several kinds of stainless steel. You can look it up on the web and I found several, minimum value of 0.490 J/g*C and maximum value of 0.530 J/g*C.
Use the same formula for stainless stell if you wish. I would use an average value of about 0.500 J/g*C, with a mass of 544 g. The delta T for the stainless steel vessel will be the same as you use for water. The vessel will be the same beginning T as the water and will be the same ending T as the water.
Then add q for H2O + q for vessel to find total q.

To calculate the amount of heat required to raise the temperature of both the water and the stainless steel vessel, we can use the equation:

Q = mcΔT

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)
ΔT is the change in temperature (in °C)

First, let's calculate the heat required to raise the temperature of the water:

m_water = 333 g
c_water = 4.18 J/g°C (specific heat capacity of water)

ΔT_water = 100°C - 16.4°C = 83.6°C

Q_water = m_water * c_water * ΔT_water
Q_water = 333 g * 4.18 J/g°C * 83.6°C

Now, let's calculate the heat required to raise the temperature of the stainless steel vessel:

m_steel = 544 g
c_steel = 0.51 J/g°C (specific heat capacity of stainless steel)

ΔT_steel = 100°C - 16.4°C = 83.6°C

Q_steel = m_steel * c_steel * ΔT_steel
Q_steel = 544 g * 0.51 J/g°C * 83.6°C

Finally, to find the total heat required, we can simply add the heat needed for the water and the steel vessel:

Q_total = Q_water + Q_steel

Now, let's calculate the values:

Q_water = 333 g * 4.18 J/g°C * 83.6°C
Q_water = 115,605 J (or approximately 115.6 kJ)

Q_steel = 544 g * 0.51 J/g°C * 83.6°C
Q_steel = 23,121.024 J (or approximately 23.1 kJ)

Q_total = 115,605 J + 23,121.024 J
Q_total = 138,726.024 J (or approximately 138.7 kJ)

Therefore, the heat that must be supplied to the stainless steel vessel containing 333 g of water is approximately 138.7 kJ.

To calculate the heat required to raise the temperature of the stainless steel vessel and water, you need to consider the specific heat capacities of stainless steel and water.

The formula to calculate the heat (q) is:
q = m × c × ΔT

where:
q = heat (in joules)
m = mass (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)

First, let's calculate the heat required to raise the temperature of the water:
m_water = 333 g
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_water = (100°C - 16.4°C) = 83.6°C

q_water = m_water × c_water × ΔT_water
q_water = 333 g × 4.18 J/g°C × 83.6°C
q_water ≈ 115,430 J

Now, let's calculate the heat required to raise the temperature of the stainless steel vessel:
m_stainless-steel = 544 g
c_stainless-steel = 0.5 J/g°C (approximate specific heat capacity of stainless steel - can vary depending on the specific alloy)
ΔT_stainless-steel = (100°C - 16.4°C)

q_stainless-steel = m_stainless-steel × c_stainless-steel × ΔT_stainless-steel
q_stainless-steel = 544 g × 0.5 J/g°C × (100°C - 16.4°C)
q_stainless-steel ≈ 41,608 J

So, the total heat required to raise the temperature of the stainless steel vessel and water is the sum of q_water and q_stainless-steel:
q_total = q_water + q_stainless-steel
q_total = 115,430 J + 41,608 J
q_total ≈ 157,038 J

Therefore, approximately 157,038 J of heat must be supplied to raise the temperature of the stainless steel vessel containing the water to the boiling point.