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

AI Thread Summary
To raise the temperature of a 13 cm³ cup of espresso from 46°C to 82°C, 4.38 grams of steam at 100°C is required. The heat needed for this temperature change is calculated using the formula Q = mcΔT, resulting in a total of 524.5 J. The volume of the steam can be determined using the ideal gas law, yielding approximately 0.11 liters. This calculation assumes steam behaves as an ideal gas under the given conditions. The solution effectively demonstrates the relationship between mass, temperature change, and volume in thermodynamic processes.
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[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.
 
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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.
 
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