calculate the heat necessary to raise the temperature of 43 grams of water from room temperature (25 C) to its boiling point

Answer 1

q = mcΔT = (43 g) (4.184 j/g ^oC) (100^oC - 25^oC) = 1.349 x 10⁴J ≅ 14 kJ

Answer 2

You will need the specific heat of water = 4.18 J/g/degree

Use the equation q = mC∆T

q = heat = ?

m = mass = 43 g

C = specific heat of water = 4.18 J/g/degree

∆T = change in temperature = 100º - 25º = 75º

q = (43 g)(4.18 J/g/degree)(75 degrees)

q = 13,490.5 J = 13,000 J (2 significant figures)