Answer:
(001) 4. Heating 100 g of water from 10°C to 50°C
(002) 4.32 km
(003) 5. B only
(004) 347°C
(005) 73.5 °C
(006) 4.9 g
(007) 3. Adding the same amount of heat to two different objects will produce the same increase in temperature.
Explanation:
(001) Heat is the product of mass, specific heat capacity, and temperature change.
q = mCΔT
The specific heat of water is 1 cal/g/°C. So the heat required for each is:
1. q = (1000 g) (1 cal/g/°C) (12°C − 10°C) = 2000 cal
2. q = (10 g) (1 cal/g/°C) (40°C − 10°C) = 300 cal
3. q = (1 g) (1 cal/g/°C) (80°C − 10°C) = 70 cal
4. q = (100 g) (1 cal/g/°C) (50°C − 10°C) = 4000 cal
(002) A "food Calorie" is 1 kcal = 1000 cal, or 4186 Joules.
Work = change in energy
W = ΔPE
W = mgΔh
758 kcal (4186 J/kcal) = (74.9 kg) (9.8 m/s²) h
h = 4320 m
h = 4.32 km
(003) Heat flows from higher temperature to lower temperature.
It's important to note that an object can have more thermal energy and still have a lower temperature. For example, the ocean has low temperature but a very large mass, so it has more thermal energy than a cup of coffee, which has higher temperature but a much smaller mass.
(004) Kinetic energy = heat
KE = q
½ mv² = mCΔT
½ v² = CΔT
ΔT = v² / (2C)
ΔT = (298 m/s)² / (2 × 128 J/kg/°C)
ΔT = 347°C
(005) Heat gained by the glass = heat lost by water
m₁C₁ΔT₁ = -m₂C₂ΔT₂
(300 g) (0.2 cal/g/°C) (T − 32°C) = -(199 g) (1 cal/g/°C) (T − 86°C)
(60 cal/°C) (T − 32°C) = -(199 cal/°C) (T − 86°C)
(60 cal/°C) T − 1920 cal = -(199 cal/°C) T + 17114 cal
(259 cal/°C) T = 19034 cal
T = 73.5 °C
(006) Heat gained by ice = heat lost by water
m₁C₁ΔT₁ = m₂L₂
(49 g) (0.5 cal/g/°C) (0°C − (-16°C)) = m (80 cal/g)
392 cal = m (80 cal/g)
m = 4.9 g
(007) Heat is the product of mass, specific heat capacity, and temperature change.
q = mCΔT
So if two objects have different masses, or different heat capacities, then adding the same amount of heat will produce different increases in temperature.
Statement 3 is false.
