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

Click For Summary
SUMMARY

The discussion focuses on calculating the grams and volume of steam required to raise the temperature of a 13 cm3 cup of espresso from 46°C to 82°C using steam at 100°C. The heat required is determined using the equation Q = mcΔT, resulting in a heat requirement of 524.5 J. This leads to the conclusion that 4.38 grams of steam is necessary. The volume of this steam, calculated using the ideal gas law (PV = nRT), is approximately 0.11 liters.

PREREQUISITES
  • Understanding of specific heat capacity (4.186 J/g°C)
  • Familiarity with the equation Q = mcΔT
  • Knowledge of the ideal gas law (PV = nRT)
  • Basic thermodynamics concepts
NEXT STEPS
  • Study the ideal gas law and its applications in thermodynamics
  • Learn about specific heat capacity and its role in heat transfer
  • Explore calculations involving phase changes and steam properties
  • Investigate the relationship between temperature, pressure, and volume in gases
USEFUL FOR

Students in physics or chemistry, engineers working with thermodynamic systems, and anyone interested in the thermal properties of steam and heat transfer calculations.

merbear
Messages
9
Reaction score
0
[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.
 
Physics news on Phys.org
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.
 

Similar threads

Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
9
Views
4K
Replies
33
Views
8K