Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
As per energy conservation in the reversible engine we can say

here we know that


now from above equation


now we can convert it into kW


so above is the power input to the refrigerator
now to find COP we know that


so COP of refrigerator is 2.2
The length of the string must vary, while the angle and acceleration are constant
True because it is the same amount and same speed but just different directions and people
<span>motion of truck constitutes of 3 travels.
1. accelerating uniformly with acceleration a1 = 2 m/s^2 until its velocity reached 20 m/s travelling a
distance of 's1' meters.
2. uniform motion with 20 m/s for a time duration t1 = 20s travelling a distance of 's2' meters.
3. uniform deceleration for t2 = 5 sec which stops the truck after travelling a distance of 's3' meters.. </span>