We are given:
Mass of ice = 100 grams
Initial temperature = -50°C
Final Temperature = -10°C
We know that the specific heat of ice is 2.09 joules / (gram * °c)
Heat Absorbed:
We know that:
ΔQ = mcΔT [where ΔQ is the heat absorbed]
ΔQ = (100 grams)(2.09)(-10 - (-50))
ΔQ = (100)(2.09)(40)
ΔQ = 8360 Joules
