An electric heater supplies 5kw of power to atank of water assume all the energy supplied is converted into heat energy and the energy losses to the surrounding are negligible how long will it take to heat 500 kg of water in the tank from 25c to 100c specifiv heat capacity water 4200j/kgc

8hours 45minutes

Explanation:

Amount of heat energy used up by the electric substance = heat gained by the water.

Amount of heat energy used up by the electric substance = Power * Time

heat gained by the water. = mcΔt

m is the mass of the water in kg

c is the specific heat capacity

Δt is the change in temperature

Given parameters

Power = 5kW = 5000W

mass of water m = 500kg

c = 4200J/kgC

Δt = 100°C - 25°C = 75°C

Required

Time taken

Substituting the given parameters into the formula

Pt = mcΔt

5000t = 500*4200*75

5000t = 157,500,000

Divide both sides by 5000

t =  157,500,000/5000

t = 31,500 secs

t = (31,500/3600)hrs

t = 8.75hours

t = 8hours 45minutes

Hence it takes 8hours 45minutes to heat 500 kg of water in the tank from 25c to 100c

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

the answer to this is proletariats.

hope this : )