1kg of ice at 0C is placed in a container with 2kg of water at 65C. The container is in an atmospheric pressure that remains at 15C. There is a free exchange of energy between all items so that a final temperature of 15C is reached. Calculate the amount of heat transferred to:
A. The original 1kg of ice
B. The water
C. The atmosphere
Indicate the direction of heat interactions in each case.
To calculate the amount of heat transferred to each component, we can use the specific heat capacity equation:
Q = mcΔT
where Q is the heat exchanged, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
For the ice:
1. First, we need to determine the heat required to raise the ice temperature from 0°C to 15°C. Since the final temperature is 15°C, there is no phase change occurring. We can use the specific heat capacity of ice, which is 2.09 J/g°C (joules per gram per degree Celsius).
Qice = (mass of ice) x (specific heat capacity of ice) x (change in temperature)
Qice = 1 kg x 1000 g/kg x 2.09 J/g°C x (15°C - 0°C)
Qice = 31,350 J
The heat transferred to the ice is +31,350 J. The positive sign indicates that heat is being gained by the ice.
For the water:
1. The water undergoes a phase change while cooling down. It goes from 65°C to 15°C, which is a change from a liquid to a liquid, so we don't need to consider any latent heat. We can use the specific heat capacity of water, which is 4.18 J/g°C.
Qwater = (mass of water) x (specific heat capacity of water) x (change in temperature)
Qwater = 2 kg x 1000 g/kg x 4.18 J/g°C x (15°C - 65°C)
Qwater = - 166,400 J
The heat transferred to the water is -166,400 J. The negative sign indicates that heat is being lost or transferred away from the water.
For the atmosphere:
1. The atmosphere doesn't undergo any change in phase, so we can simply use the specific heat capacity of air, which is approximately 1.01 J/g°C.
Qatmosphere = (mass of air) x (specific heat capacity of air) x (change in temperature)
To calculate the mass of air, we need to know the volume. Let's assume the volume of the container is 1 liter or 1000 cm^3.
Density of air at 15°C and 1 atm pressure = 1.184 kg/m^3.
Volume of air = 1000 cm^3 = 1/1000 m^3 = 0.001 m^3.
Mass of air = (density of air) x (volume of air)
Mass of air = 1.184 kg/m^3 x 0.001 m^3 = 0.001184 kg.
Qatmosphere = 0.001184 kg x 1000 g/kg x 1.01 J/g°C x (15°C - 15°C)
Qatmosphere = 0 J
The heat transferred to the atmosphere is 0 J, indicating no heat exchange with the atmosphere.
So, the answers are:
A. The original 1 kg of ice gains 31,350 J of heat. (+31,350 J)
B. The water loses 166,400 J of heat. (-166,400 J)
C. There is no heat exchange with the atmosphere. (0 J)
heat to the ice: 1kg*Hfusion
heat to water? Which water, the melted ice? the original water?
c. atmospere: zero