Water (2410 g) is heated until it just begins to boil. If the water absorbs 5.01×105J of heat in the process, what was the initial temperature of the water?

Question

Express your answer numerically in degrees Celsius.

Answer 1

Hi,

For this question, use the equation q = mcΔT

q = heat (J)

m = mass (g)

c = specific heat ( for water = 4.184 J/g °C)

Δ T = T final - T initial (Tf - Ti)

Since the water is just beginning to boil, it's safe to assume that the temperature is 100°C, which is the b.p. of water.

Plugging in the given and known values:

5.01 x 10⁵ J = 2410 g x 4.184 J/g °C x (Tf - Ti)

5.01 x 10⁵ J = 10083.44 J/°C x (Tf - Ti)

49.68 °C = 100°C - Ti

And then solve for Ti

Hope this helps!