Question

50 Points!

Answer 1

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 = $T(i) = -22\°C$

Final temperature of steam = $T(f) = 118\°C$

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) = $T(f)-T(i)$

Specific heat of water is c = 4.186 J /g. °C

Substituting these in equation (1), we get

Q = m×c×Δt = $m\times c\times (T(f)-T(i))$

$Q = 10.75\times 4.186\times (118-(-22))\\ Q= 10.75\times 4.186\times 140 = 6299.93\ J$

Required heat energy is equal to 6299.93 Joules ≅ 6300 Joules.