Question

The specific heat of water in its solid phase (ice) is 2090 J/(kg K), while in the liquid phase (water) its specific heat is 4190 J/(kg K). Water's latent heat of fusion is 333,000 J/kg. If you have a 2kg block of ice at -90°C and you add 1,000,000 J of heat, what is its new temperature?

a. 0°C
b. 14°C
c. 49°C
d. 149°C

Answer 1

d. 149 ⁰C.

Step-by-step explanation:

Given;

mass of the block of ice, m = 2 kg

specific heat capacity of the ice, C = 2090 J/(kgK)

initial temperature of the ice, t₁ = -90 ⁰C

heat added to the ice, H = 1,000,000 J

let the final temperature of the ice = t₂

The final temperature of the ice after adding the heat is calculated as follows;

`H = mC_(ice) \Delta t\\\\H = mC_(ice) (t_2 - t_1)\\\\1,000,000 = 2 * 2090 * (t_2 - (-90))\\\\1,000,000 = 4,180(t_2 +90)\\\\1,000,000 = 4,180t_2 + 376,200\\\\1,000,000 - 376,200 = 4,180t_2\\\\623,800 = 4,180 t_2\\\\t_2 = (623,800)/(4,180) \\\\t_2 = 149 \ ^0C`

Therefore, the new temperature of the water is 149 ⁰C.