Grams and volume of steam needed to raise the temperature of expresso

[merbear]

[SOLVED] Grams and volume of steam needed to raise the temperature of expresso

Homework Statement

The temperature of espresso coffee (mostly water) can be increased by blowing 100°C steam into it. How much steam (in grams) is needed to heat up a 13 cm3 cup of espresso from 46°C to 82°C? What is the volume of this quantity of steam, assuming that the steam is an ideal gas

Homework Equations

Q=mcdeltaT

dq=Cv*dt

The Attempt at a Solution

I know that the specific heat- amount of heat needed to raise the temperature 1 degree -is 4.186 J/gC. But I really don't have any idea as to how to start the problem. I tried using Q=mcdeltaT, but I don't have a value for Q.

I would appreciate any help in getting started. Thank you.

 

[Nate810]

Solution:We can use the equation Q = mcΔT to solve for the mass of steam needed. The heat required to raise the temperature from 46°C to 82°C is given by Q = (13 cm3) (4.186 J/g°C) (82°C - 46°C) = 524.5 J. Thus, the mass of steam needed is m = 524.5 J / (100°C - 46°C) (4.186 J/g°C) = 4.38 g.The volume of this quantity of steam can be calculated using the ideal gas law, PV = nRT, where n is the number of moles of steam and R is the ideal gas constant, which has a value of 8.314 J/molK. We can rearrange the equation to calculate the volume of steam as V = nRT/P = (4.38 g)(8.314 J/molK)(273.15 K)/(1.01325 x 105 Pa) = 0.11 L.

 

[Moetasim]

To solve this problem, we first need to find the amount of heat (Q) needed to raise the temperature of the espresso from 46°C to 82°C. This can be done using the equation Q=mcdeltaT, where m is the mass of the espresso, c is the specific heat of water (4.186 J/gC), and deltaT is the change in temperature (82°C-46°C = 36°C).

Since we are given the volume of the espresso (13 cm3), we can use the density of water (1 g/cm3) to find the mass of the espresso.

m = density*volume
= (1 g/cm3)*(13 cm3)
= 13 g

Now, we can plug in these values into the equation Q=mcdeltaT:

Q = (13 g)*(4.186 J/gC)*(36°C)
= 1901.64 J

This is the amount of heat needed to raise the temperature of the espresso.

Next, we need to find the amount of heat (Q') released by the steam as it condenses. This can be done using the equation dq=Cv*dt, where Cv is the specific heat of steam at constant volume (1.84 J/gC) and dt is the change in temperature of the steam (100°C-82°C = 18°C).

Q' = Cv*dt
= (1.84 J/gC)*(18°C)
= 33.12 J

Since the heat released by the steam needs to be equal to the heat absorbed by the espresso, we can set Q=Q' and solve for the mass of steam (m').

Q = Q'

mcdeltaT = Cv*dt

m' = (Cv*dt)/(c*deltaT)
= (1.84 J/gC)*(18°C)/(4.186 J/gC)*(36°C)
= 0.244 g

Therefore, we need 0.244 grams of steam to raise the temperature of the espresso from 46°C to 82°C.

To find the volume of this quantity of steam, we can use the ideal gas law:

PV = nRT

where P is the pressure of the steam (atmospheric pressure), V is the volume of the steam, n is the number of moles of steam, R is the ideal gas constant (0.