30.1 grams of ice at -250^oC would be required how many kj to increase it's temperature to 15.0^ᴏC

16.1923 kj

Explanation:

Given data:

Mass of ice = 30.1 g

Initial temperature = -250 °C

Final temperature = 15.0 °C

Energy required = ?

Solution:

Specific heat capacity of ice is 2.03 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT  = 15.0°C - (-250 °C)

ΔT  = 265°C

Q = 30.1 g × 2.03 J/g.°C × 265°C

Q = 16192.3 J

J to kJ:

16192.3 J × 1 KJ /1000 j

16.1923 kj

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.