Respuesta :
A heat energy equal to 6300 J is required to heat 10.75 g of ice to steam.
Explanation:
Given:
Amount of transferred energy = Q =?
Mass of ice (water) = m = 10.75 g
Initial temperature of ice = [tex]T(i) = -22\°C[/tex]
Final temperature of steam = [tex]T(f) = 118\°C[/tex]
Formula for Heat capacity is given by
Q = m×c×Δt ........................................(1)
where:
Q = Heat capacity of the substance (in J)
m=mass of the substance being heated in grams(g)
c = the specific heat of the substance in J/(g.°C)
Δt = Change in temperature (in °C)
Δt = (Final temperature - Initial temperature) = [tex]T(f)-T(i)[/tex]
Specific heat of water is c = 4.186 J /g. °C
Substituting these in equation (1), we get
Q = m×c×Δt = [tex]m\times c\times (T(f)-T(i))[/tex]
[tex]Q = 10.75\times 4.186\times (118-(-22))\\ Q= 10.75\times 4.186\times 140 = 6299.93\ J[/tex]
Required heat energy is equal to 6299.93 Joules ≅ 6300 Joules.
