Explanation:
(1) The heat added to warm the ice to 0°C is:
q = mCΔT = (0.041 kg) (2090 J/kg/°C) (0°C − (-11°C)) = 942.59 J
The heat added to melt the ice is:
q = mL = (0.041 kg) (3.33×10⁵ J/kg) = 13,653 J
The heat added to warm the water to 100°C is:
q = mCΔT = (0.041 kg) (4186 J/kg/°C) (100°C − 0°C) = 17,162.6 J
The heat added to evaporate the water is:
q = mL = (0.041 kg) (2.26×10⁶ J/kg) = 92,660 J
The heat added to warm the steam to 115°C is:
q = mCΔT = (0.041 kg) (2010 J/kg/°C) (115°C − 100°C) = 1236.15 J
The total heat needed is:
q = 942.59 J + 13,653 J + 17,162.6 J + 92,660 J + 1236.15 J
q = 125,654.34 J
(2) When the first two are mixed:
m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0
C₁ (T₁ − T) + C₂ (T₂ − T) = 0
C₁ (6 − 11) + C₂ (25 − 11) = 0
-5 C₁ + 14 C₂ = 0
C₁ = 2.8 C₂
When the second and third are mixed:
m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0
C₂ (T₂ − T) + C₃ (T₃ − T) = 0
C₂ (25 − 33) + C₃ (37 − 33) = 0
-8 C₂ + 4 C₃ = 0
C₂ = 0.5 C₃
Substituting:
C₁ = 2.8 (0.5 C₃)
C₁ = 1.4 C₃
When the first and third are mixed:
m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0
C₁ (T₁ − T) + C₃ (T₃ − T) = 0
(1.4 C₃) (6 − T) + C₃ (37 − T) = 0
(1.4) (6 − T) + 37 − T = 0
8.4 − 1.4T + 37 − T = 0
2.4T = 45.4
T = 18.9°C
(3) Heat gained by the ice = heat lost by the tea
mL + mCΔT = -mCΔT
m (3.33×10⁵ J/kg) + m (2090 J/kg/°C) (30.8°C − 0°C) = -(0.176 kg) (4186 J/kg/°C) (30.8°C − 32.8°C)
m (397372 J/kg) = 1473.472 J
m = 0.004 kg
m = 4 g
4 grams of ice is melted and warmed to the final temperature, which leaves 128 grams unmelted.
(4) The heat added to warm the ice to 0°C is:
q = mCΔT = (0.028 kg) (2090 J/kg/°C) (0°C − (-67°C)) = 3920.84 J
The heat added to melt the ice is:
q = mL = (0.028 kg) (3.33×10⁵ J/kg) = 9324 J
The heat added to warm the melted ice to T is:
q = mCΔT = (0.028 kg) (4186 J/kg/°C) (T − 0°C) = (117.208 J/°C) T
The heat removed to cool the water to T is:
q = -mCΔT = -(0.505 kg) (4186 J/kg/°C) (T − 27°C)
q = (2113.93 J/°C) (27°C − T) = 57076.11 J − (2113.93 J/°C) T
The heat removed to cool the copper to T is:
q = -mCΔT = -(0.092 kg) (387 J/kg/°C) (T − 27°C)
q = (35.604 J/°C) (27°C − T) = 961.308 J − (35.604 J/°C) T
Therefore:
3920.84 J + 9324 J + (117.208 J/°C) T = 57076.11 J − (2113.93 J/°C) T + 961.308 J − (35.604 J/°C) T
13244.84 J + (117.208 J/°C) T = 58037.418 J − (2149.534 J/°C) T
(2266.742 J/°C) T = 44792.58 J
T = 19.8°C
(5) Kinetic energy of the hammer = heat absorbed by ice
KE = q
½ mv² = mL
½ (0.8 kg) (0.9 m/s)² = m (80 cal/g × 4.186 J/cal × 1000 g/kg)
m = 9.68×10⁻⁷ kg
m = 9.68×10⁻⁴ g
(6) Heat rate = thermal conductivity × area × temperature difference / thickness
q' = kAΔT / t
q' = (1.09 W/m/°C) (4.5 m × 9 m) (10°C − 4°C) / (0.09 m)
q' = 2943 W
After 10.7 hours, the amount of heat transferred is:
q = (2943 J/s) (10.7 h × 3600 s/h)
q = 1.13×10⁸ J
q = 113 MJ