Answer:
Q = 131670 ×10⁻⁶ MJ
Explanation:
Given data:
Mass of ice = 7 Kg
Initial temperature = -9 °C ( -9+273 = 264 k)
Final temperature = 0°C (0 +273 = 273 k)
Heat absorbed = ?
Solution:
Q = m. c . ΔT
Specific heat capacity of ice is 2090 j/Kg.K
ΔT = T2 - T1
ΔT = 273 k - 264 k
ΔT = 9 k
Q = m. c . ΔT
Q = 7kg. 2090 j/Kg.K . 9 k
Q = 131670 j
Q = 131670 / 10⁶ MJ
Q = 131670 ×10⁻⁶ MJ
