Answer:
4200 J/°C/kg
Explanation:
The formula for heat transfer is given by :
Q= m*c*ΔT where;
Q= heat transferred = 6930 J
m=mass of the liquid = 0.330 kg
c= specific heat capacity=?
ΔT = 25-20 = 5.0°C
Applying the values in the formula as;
Q= m*c*ΔT
6930 = 0.330 * c * 5
6930 = 1.65 c
6930/1.65 = c
4200 = c
c= 4200 J/°C/kg
32.6% of 1.4 Mw
= (0.326) x (1,400,000 joules/second)
= 456,400 joules/second .
1 year
= (365 da) x (86,400 sec/da) = 31,536,000 seconds
(456,000 joules/sec) x (31,536,000 sec) = 1.438 x 10¹³ Joules
(That's 1.438 x 10⁷ Megajoules, or 3,994,560 kWh)
Answer:
13330.86 J
Explanation:
mass of volume of 30.1 cc of water = density x volume
= 1 x 30.1 = 30.1 gm
= 30.1 x 10⁻³ kg
Heat to be withdrawn to cool water from 22.4 degree to 0 degree
= mass x specific heat x fall of temperature
= 30.1 x 10⁻³ x 4186 x 22.4
= 2822.36 J
Heat to be withdrawn to cool water 0 degree ice
mass x latent heat of freezing
30.1 x 10⁻³ x 334000
= 10053.4 J
Heat to be withdrawn to cool ice from 0 degree to -7.2 degree
mass x specific heat of ice x fall of temperature
= 30.1 x 10⁻³ x 2100 x 7.2
= 455.1 J
Total heat to be withdrawn
= 2822.36 + 10053.4 + 455.1
= 13330.86 J
Answer:
m=6.25kg
Explanation:
Force= mass × acceleration
50=m×8
make m subject of the formula...
m=50/8
m=6.25kg