Question

How much energy is need to turn 48000g of ice at -25 degrees celsius into steam at 110 degrees celsius. Don’t forget units and sig figs—also use scientific notation.

Answer 1

The specific heat capacity of ice is 2.092 J/g°C, the specific heat capacity of water is 4.184 J/g°C, and the specific heat capacity of steam is 2.010 J/g°C. The latent heat of fusion of water is 333.55 J/g, and the latent heat of vaporization of water is 2257 J/g.

The total energy required to turn 48000g of ice at -25°C into steam at 110°C is:

(48000 g)(2.092 J/g°C)(25°C) + (48000 g)(4.184 J/g°C)(85°C) + (48000 g)(333.55 J/g) + (48000 g)(2257 J/g)

= 26462400 J

= 2.646 × 10^6 J

To express the answer in scientific notation with 3 significant figures, we can write:

E = 2.65 × 10^6 J