Ch 16 Problem Set 4:
Heat transfer in Liquids and Solids KEY
Equations and constants:
q = mct (for problems involving changes in temperature)
q = m·Hfus or q = m·Hvap (for problems involving phase changes)
Specific heat of ice = 2.09 J/g·ºC
Specific heat of water = 4.18 J/g·ºC
Specific heat of steam = 2.03 J/g·ºC
Heat if fusion of water = 334 J/g
1.
How much heat must your body transfer to 500.0g of water to heat it from 25.0ºC to body
temperature, 37.0ºC?
q = mcΔt
q = (500.0 g)( 4.18 J/g•°C)(8.0 °C)
q = 25,080 J ≈ 25.1 kJ
2. How much heat energy from the sun is needed to heat a 1575g puddle of water from 5.00ºC in the
morning to 20.0ºC by the afternoon?
q = mcΔt
q = (1575 g)(4.18 J/g•°C)(15.0 °C)
q = 98,753 J ≈ 98.8 kJ
3. How much heat is needed to warm a 50.0g piece of solid copper from 25.0ºC to 200.0ºC? (the
specific heat of copper is 0.385 J/g·ºC)
q = mcΔt
q = (50.0 g)(0.385 J/g•°C)(175.0 °C)
q = 3,369 J ≈ 3.37 kJ
4. How much energy is needed to heat a 35.5g sample of ice at -17.5ºC to liquid water at 77.3ºC?
q-17.5→0 = (35.5 g)(2.09 J/g•°C)(17.5 °C) = 1,298 J = 1.30 kJ
qmelting = (35.5 g)(344 J/g•°C) = 11,857 J = 11.9 kJ
q0→77.3 = (35.5 g)(4.18 J/g•°C)(77.3 °C) = 11,471 J = 11.5 kJ
qtotal = 24.7 kJ
5. How much energy is needed to heat a 68.9g sample of water at 88.5ºC to steam at 103.7ºC?
q88.5→100 = (68.9 g)( 4.18 J/g•°C)(11.5 °C) = 3312J = 3.31 kJ
qvaporization = (68.9 g)(2260 J/g)= 155,714 J = 155 kJ
q100→103.7 = (68.9 g)(2.03 J/g•°C)(3.7 °C) = 518 J = .518 kJ
qtotal = 159 kJ
6. How much heat energy is released when a 234.7g sample of steam at 114.5ºC is cooled until it’s ice
at a temperature of -8.50ºC?
q114.5100 = (234.7 g)(2.03 J/g•°C)(-14.5 °C) = -6,908 J = -6.91 kJ
qcondensation = (234.7 g)(-2260 J/g)= -530,422 J = -530. kJ
q1000 = (234.7 g)(4.18J/g•°C)(-100.0 °C) = -98,105 J = -98.1 kJ
qfreezing = (234.7 g)(-334 J/g) = -78,390. J = -78.4 kJ
q0-8.5 = (234.7 g)( 2.09 J/g•°C)(-8.5 °C) = -4,169 J = -4.17 kJ
qtotal = 718 kJ
Heats of Formation
#7-10 Use the standard heats of formation
7) Br2 (g)  Br2 (l) (values for this prob. not in Honors Chem text – use handout)
Hp - Hr = 0 – (+30.91 kJ/mol) = -30.91 kJ/mol
8) CaCO3  CaO + CO2
[(-635.1) + (-393.5)] – (-1206.9) = 178.3 kJ/mol
9) 2NO(g) + O2  2NO2
[2(33.2)] – [2(90.2) + 0] = 66.4 – 180.4 = -114.0 kJ/mol
10) 4NH3 + 5O2  2NO + 6H2O(l)
[6(-285.8) + 2(90.2)] – [4(-46.1) + 0] = -1714.8 + 180.4 + 184.4 = -1350.0 kJ/mol