Answer:
Q = 2260.44 j
Explanation:
Given data:
Mass of ice = 18 g
Initial temperature = -10 °C
Final temperature = 20°C
Heat absorbed by ice = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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 = Final temperature - initial temperature
ΔT = 20°C - (-10°C)
ΔT = 30°C
Specific heat of water = 2.00 j/g.°C
Q = 18 g × 4.186 j/g.°C × 30°C
Q = 2260.44 j
