Answer:
1.07 ×10² j/g.°C
Explanation:
Given data:
Mass of substance Z = 13.5 g
Amount of heat absorbed = 36 Kj (36×1000 = 36000 j)
Increase in temperature= ΔT = 25°C
Specific heat of substance = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values in formula.
Q = m.c. ΔT
36000 j = 13.5 g ×c×25°C
36000 j / 13.5 g×25°C = c
c = 36000 j /337.5 g.°C
c = 1.07 ×10² j/g.°C
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Answer:
K.E = 19125 j
Explanation:
Given data:
Mass = 150 Kg
Speed of car = 15 m/s
K.E = ?
Solution:
Formula:
K.E = 1/2 (mv²)
K.E = Kinetic energy
m = given mass
V = speed
now we will put the values in formula.
K.E = 1/2 (mv²)
K.E = 1/2 (150 Kg) (15²)
K.E = 1/2 150 Kg × 225 m²/s²
K.E = 0.5 × 225 m²/s² ×150 Kg
K.E = 19125 j
