Answer:
16.1923 kj
Step-by-step 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