Calorimetry Practice Problems (use q = mcat) c for water is 4.184 J/g degree C a

Question

Calorimetry Practice Problems (use q = mcat) c for water is 4.184 J/g degree C and c for steam is 2 08 J/g degree C How much energy must be absorbed by 20.0 g of steam to increase its temperature from 283.0 degree C to 303.0 degree C? How much energy is required to heat 120.0g of water from 2.0 degree C to 24.0 degree C? If 720.0 g of steam at 400.0 degree C absorbs 800.0 kJ of heat energy, what will be its increase in temperature? How much heat (in kJ) is given out when 85.0 g of lead cools from 200.0 degree C to 10.0 degree C? (C_p of lead = 0.129 J/g degree C) If it takes 41.72 joules to heat a piece of gold weighing 18.69 g from 10.0 degree C to 27.0 degree C, what is the specific heat of the gold? A certain mass of water was heated with 41, 840 Joules, raising its temperature from 22.0 degree C to 28.5 degree C. Find the mass of water. Perform the following conversions. Use scientific notation when appropriate. (4.184 J = 1 cal)

Explanation / Answer

Q12.

Qsteam = mass of steam * Cp steam * (Tf-Ti)

Q = 20 g *2.08 J/gC *(303-283)°c

Q = 832 J required

Q13.

Energy for water :

Q = m*C*(Tf-Ti)

Q = 120 g *4.184 J/gC *(24-2)C

q = 11045.76 J

Q14

Find dT:

Qsteam = mass of steam * Cp steam * (Tf-Ti)

Tf-Ti = Qsteam /(Qsteam = mass of steam * Cp steam * (Tf-Ti))

dT = (800*10^3)/(720*2.08) = 534.188 °C

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