The amount of heat Q needed to turn a mass m of room temperature T_1 water into steam at 100degrees C T_2 can be found using the specific heat c of water and the heat of vaporization H_v of water at 1 atmosphere of pressure.

Suppose that a commercial espresso machine in a coffee shop turns 1.50 kg of water at 22.0degreesC into steam at 100degreesC. If c = 4187 and H_v=2258 kJ/kg, how much heat Q is absorbed by the water from the heating resistor inside the machine?
Assume that this is a closed and isolated system.

q1 = heat needed to heat 1.5 kg water from room T to 100 C.

q1 = mass x specific heat x (Tfinal-Tinitial).
q1 = 1.5 kg x 4187 J/kg*C x (100-22) = ??

q2 = heat needed to turn 1.5 kg water at 100 C to steam at 100 C = mass x Hvap
q2 = 1.5 kg x 2,258,000 J/kg = ??

Total heat is q1 + q2.
Check my thinking.

To calculate the amount of heat Q absorbed by the water from the heating resistor inside the espresso machine, we will use the formula:

Q = mcΔT + mL

Where:
Q is the amount of heat absorbed
m is the mass of water in kilograms
c is the specific heat capacity of water in J/kg°C
ΔT is the change in temperature in °C
L is the heat of vaporization of water at a given pressure in J/kg

Given values:
m = 1.50 kg
c = 4187 J/kg°C
T₁ = 22.0°C
T₂ = 100°C
Hᵥ = 2258 kJ/kg

First, let's convert the heat of vaporization from kilojoules to joules:
Hᵥ = 2258 kJ/kg = 2258 × 1000 J/kg = 2,258,000 J/kg

Now, let's calculate the amount of heat absorbed:

ΔT = T₂ - T₁ = 100°C - 22.0°C = 78.0°C

Q = (1.50 kg)(4187 J/kg°C)(78.0°C) + (1.50 kg)(2,258,000 J/kg)
Q = 474,645 J + 3,387,000 J
Q = 3,861,645 J

Therefore, the amount of heat Q absorbed by the water from the heating resistor inside the machine is 3,861,645 Joules.

To find the amount of heat Q absorbed by the water from the heating resistor inside the machine, we can use the equation:

Q = (m * c * ΔT) + (m * Hv)

Where:
Q is the amount of heat absorbed by the water
m is the mass of water
c is the specific heat of water
ΔT is the change in temperature (T2 - T1)
Hv is the heat of vaporization of water

Given:
m = 1.50 kg
c = 4187 J/kg°C (since q is given in J)
ΔT = 100°C - 22.0°C
Hv = 2258 kJ/kg

First, let's convert Hv to J/kg:
Hv = 2258 kJ/kg * 1000 J/1 kJ = 2258000 J/kg

Now let's substitute the values into the equation:

Q = (1.50 kg * 4187 J/kg°C * (100°C - 22.0°C)) + (1.50 kg * 2258000 J/kg)

Simplifying the equation:

Q = (1.50 kg * 4187 J/kg°C * 78°C) + (1.50 kg * 2258000 J/kg)

Q = 388194 J + 3387000 J

Q = 3775194 J

Therefore, the amount of heat Q absorbed by the water from the heating resistor inside the machine is 3775194 J.