Answer:
Latent heatnof fusion = 417.5 J
Explanation:
Specific latent heat of fusion of water is 334kJ.kg-1.
The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.
The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.
Latent heat = ML
Latent heat= 1.25 kg * 334kJ.kg-1
Latent heat = 1.25*334 *(J/kg)*kg
Latent heat = 417.5 J
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
All of the following are non-renewable resources except
O natural gas
O oil
O minerals
O <em>water ✓ </em>
- <em>Water </em><em>is </em><em>a </em><em>renewable </em><em>source </em><em>because </em><em>evaporation </em><em>and </em><em>condensation </em><em>takes </em><em>place </em><em>everytime </em><em>on </em><em>our </em><em>planet</em>
