Answer:
increase in temperature of water is 10° C
Explanation:
Given data
pizza = 500 kcal = 500000 calories
cold water = 50 L
to find out
increase in temperature of water
solution
we know heat formula that is
heat = m × specific heat × Δt
here m is mass = 50000 gram and Δt is change in temperature
and specific heat = 1 cal / gram C
so put here all value and find Δt
500000 = 50000 × 1 × Δt
Δt = 10° C
so increase in temperature of water is 10° C
Answer:

& 
Explanation:
Given:
- interior temperature of box,

- height of the walls of box,

- thickness of each layer of bi-layered plywood,

- thermal conductivity of plywood,

- thickness of sandwiched Styrofoam,

- thermal conductivity of Styrofoam,

- exterior temperature,

<u>From the Fourier's law of conduction:</u>

....................................(1)
<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>




.....................(2)
Putting the value from (2) into (1):


is the heat per unit area of the wall.
The heat flux remains constant because the area is constant.
<u>For plywood-Styrofoam interface from inside:</u>



&<u>For Styrofoam-plywood interface from inside:</u>



Answer:
One characteristic of mass wasting processes is that they move materials relatively short distances compared to streams. - d.
Answer:
(A) power = 0.208 kW = 208 watts
(B) energy = 6.6 x 10^{9} joules
Explanation:
energy consumed per day = 5 kWh
(a) find the power consumed in a day
1 day = 24 hours
power = \frac{energy}{time}
power = \frac{5}{24}
power = 0.208 kW = 208 watts
(b) find the energy consumed in a year
assuming it is not a leap year and number of days = 365 days
1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds
energy = power x time
energy = 208 x 31,536,000
energy = 6.6 x 10^{9} joules
Answer:
The change in kinetic energy (KE) of the car is more in the second case.
Explanation:
Let the mass of the car = m
initial velocity of the first case, u = 22 km/h = 6.11 m/s
final velocity of the first case, v = 32 km/h = 8.89 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(8.89² - 6.11²)
= 20.85m J
initial velocity of the second case, u = 32 km/h = 8.89 m/s
final velocity of the second case, v = 42 km/h = 11.67 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(11.67² - 8.89²)
= 28.58m J
The change in kinetic energy (KE) of the car is more in the second case.