9086.3J
Explanation:
Given parameters:
Mass of water = 18.5g
Initial temperature of water = 35°C
Finial temperature = -5°C
Unknown:
Amount of heat given off = ?
Solution:
In converting the water to ice, the water goes through these processes:
35°C to freezing point at 0°C = mcΔT
0°C of water to 0°C of ice = mL
0°C of ice to -5°C = mcΔT
The heat given off;
H = mc₁ΔT + mL + mc₂ΔT
m is the mass of water
c₁ is the specific heat capacity of water = 4.2J/g °C
c₂ is the specific heat capacity of ice = 2.03J/g °C
L is the latent heat of fusion = 334J/g
ΔT = T₂ - T₁ = temperature change
H = 18.5 x 4.2 + (0-(-35)) + (18.5 x 334) + (18.5 x 2.03 x (-5 - (0))
H = 2719.5J + 6179J + 187.8J = 9086.3J
learn more:
Specific heat brainly.com/question/7210400
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