Answer:
Heat energy needed = 1243.45 kJ
Explanation:
We have
heat of fusion of water = 334 J/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C
Here wee need to convert 0.500 kg water from 50°C to vapor at 110°C
First the water changes to 100°C from 50°C , then it changes to steam and then its temperature increases from 100°C to 110°C.
Mass of water = 500 g
Heat energy required to change water temperature from 50°C to 100°C
![H_1=mC\Delta T=500\times 4.18\times (100-50)=104.5kJ](https://tex.z-dn.net/?f=H_1%3DmC%5CDelta%20T%3D500%5Ctimes%204.18%5Ctimes%20%28100-50%29%3D104.5kJ)
Heat energy required to change water from 100°C to steam at 100°C
![H_2=mL=500\times 2257=1128.5kJ](https://tex.z-dn.net/?f=H_2%3DmL%3D500%5Ctimes%202257%3D1128.5kJ)
Heat energy required to change steam temperature from 100°C to 110°C
![H_3=mC\Delta T=500\times 2.09\times (110-100)=10.45kJ](https://tex.z-dn.net/?f=H_3%3DmC%5CDelta%20T%3D500%5Ctimes%202.09%5Ctimes%20%28110-100%29%3D10.45kJ)
Total heat energy required
![H=H_1+H_2+H_3=104.5+1128.5+10.45=1243.45kJ](https://tex.z-dn.net/?f=H%3DH_1%2BH_2%2BH_3%3D104.5%2B1128.5%2B10.45%3D1243.45kJ)
Heat energy needed = 1243.45 kJ
Answer:
W = 7591.56 J
Explanation:
given,
distance of the box, d = 37 m
Force for pulling the box, F = 217 N
angle of inclination with horizontal,θ = 19°
We know,
Work done is equal to product of force and the displacement.
W = F.d cos θ
W = 217 x 37 x cos 19°
W = 7591.56 J
Hence, the work done to pull the box is equal to W = 7591.56 J
Answer:
Speed of the car after the collision is 1.2 m/s
Explanation:
It is given that,
Mass of first car, m₁ = 2 kg
Velocity of first car, v₁ = 6 m/s
Mass of second car, m₂ = 3 kg
Velocity of second car, v₂ = -2 m/s (it is travelling in opposite direction)
The two cars lock together and move along the road. This shows an inelastic collision. Let their common velocity is V. On applying the conservation of momentum as :
![m_1v_1+m_2v_2=(m_1+m_2)V](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%3D%28m_1%2Bm_2%29V)
![V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bm_1v_1%2Bm_2v_2%7D%7B%28m_1%2Bm_2%29%7D)
![V=\dfrac{2\ kg\times 6\ m/s+3\ kg\times (-2\ m/s)}{5\ kg}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B2%5C%20kg%5Ctimes%206%5C%20m%2Fs%2B3%5C%20kg%5Ctimes%20%28-2%5C%20m%2Fs%29%7D%7B5%5C%20kg%7D)
V = 1.2 m/s
So, after collision the speed of the cars is 1.2 m/s. Hence, this is the required solution.