How much heat must be removed to completely freeze 15 kg of water initially at 95°F?...
When 15-kg of some unknown at 180°C is added to 4-kg of water at 20°C the resulting temperature is 60°C. What is the specific heat of the unknown? J Cwater = 4186 kg:K, Lf = 3 For water : Cice = 2100 kg.k - 34 x 105 kg
How much ice is in the final state after 2 kg of ice at -30°C is added to a large quantity of water at 0°C? The final state is ice-water. (The temperature of the ice increases to 0°C and some of the water freezes.) J For water : Cice = 4186 J kg.K If = 3 = 2100 kg.K : Cwater L, = 2.26 x 106 .34 x 10% J kg 9 kg
How much heat must be removed from 456 g of water at 25.0 degree C to change it into ice at -10.0 degree C? The specific heat of ice is 2090 J/kg K. the latent heat of fusion of water is 33.5 times 10^4 J/Kg, and the specific heat of water is 4186 J/kg K.
How much heat is required to convert solid ice with a mass of 850 g and at a temperature of -26.5 °C to liquid water at a temperature of 64.5 °C? (The specific heat of ice is cice = 2100 J/kgK, the specific heat of water is cwater = 4186.8 J/kgK, and the heat of fusion for water is: Lf = 334 kJ/kg.)
How much heat is required to convert solid ice with a mass of 585 g and at a temperature of -29.0 °C to liquid water at a temperature of 74.5 °C? (The specific heat of ice is cice = 2100 J/kgK, the specific heat of water is cwater = 4186.8 J/kgK, and the heat of fusion for water is: Lf = 334 kJ/kg.)
When 1.4 × 105J of energy is removed from 0.9 kg of water initially at 20 ∘C will all the water freeze? The heat of fusion for water at 0 ∘C is Lf =3.35×105J/kg. The specific heat of water is c = 4180 J/kg⋅ ∘C. How much remains unfrozen?
How much heat is required to convert solid ice with a mass of 525 g and at a temperature of-23.5 °C to liquid water at a temperature of 48.5 °C? (The specific heat of ice is cice = 2100 /kgK, the specific heat of water is cwater-4186.8 J/kgK, and the heat of fusion for water is: Lf 334 kJ/kg.) Submit Answer Tries 0/12
How much heat must be absorbed by 20 g of ice at minus 16 oC to transform into 20 g of water at 45 oC. (cice= 2.05 kJ/kg, cwater=4.18 kJ*K/kg, csteam=2.01 kJ*K/kg; Lfwater=333.5 kJ/kg @ 0 oC, Lvwater= 2257 kJ/kg @100 oC) Q= J
A beaker with negligible mass contains 0.50 kg of water at a temperature of 80 °C. How many grams of ice at a temperature of -20 °C must be dropped in the water so that the final temperature of of the system is 30 °C? (Cice = 2100 J/ (kg °C) ), Cwater = 4190 J/ (kg °C) ), LF = 3.33 x 105 J/kg Could you also show how LF is used?
How much heat energy in joules must be added to 78 grams of water initially at 23 degrees Celsius if it ends up at 53 degrees Celsius. Remember that the specific heat capacity of water = 4186 J/ (kg oC) or 1 calorie / (g oC)