Respuesta :
Heat require to boil 750 g water from 25 C0to 100 C0
Q = m s dT
= 750 g * 4.2 cal /g /K * (100 - 25)
= 236.25 * 103 J
If stove is supplying 1000 W power then time required is:
Q = P *t
236.25 * 103 = 1000 * t
t = 236.25 s
Thus 236.25 s is required by a 1000 W power supplying stove to boil 750 mL water
Heat require to boil 750 g water can be calculated from the formula:
[tex] Q=m\times s\times dT [/tex]
Here, Q is the heat
m, is the mass
s is the specific heat
dT is the temperature difference
Here ,
m, is 750 g
s of water is 4.2 cal g⁻¹ K⁻
dT , as water is boiled from 25 to 100 C
Q=[tex] 750\times 4.2\times (100 - 25) [/tex]
= 236250 J
Assuming if input power is 1000 W then time required is:
Q=[tex] P\times t [/tex]
[tex] 236250 = 1000 \times t [/tex]
t = 236.25 s
Thus time required to boil 750 g of water will be 236.25 s
