Abath contains 100kg of water at 60°C.Hot and cold taps are then turned on to deliver 20kg per minute each at temperatures of 70°C and 10°C respectively. How long will it be before the temperatures in the bath have dropped to 45°C.Assume complete mixing of the water and ignore heat loses.
∑ (mass x specific heat capacity x temp change) = 0
100*c*(45-60)+20*time*c*(45-70)+20(time)*c*(45-10)=0 solve for time
To solve this problem, we need to calculate the amount of heat gained or lost by the water in the bath during the mixing process.
The specific heat capacity of water is 4.18 J/g°C.
1. Calculate the initial heat content of the water in the bath:
Initial heat content = mass × specific heat capacity × temperature
Initial heat content = 100 kg × 4.18 J/g°C × 60°C
2. Calculate the heat added by the hot water:
Heat added by hot water = mass of hot water × specific heat capacity × change in temperature
Heat added by hot water = 20 kg/min × 60 min × 4.18 J/g°C × (70°C - 45°C)
3. Calculate the heat removed by the cold water:
Heat removed by cold water = mass of cold water × specific heat capacity × change in temperature
Heat removed by cold water = 20 kg/min × 60 min × 4.18 J/g°C × (45°C - 10°C)
4. Calculate the net change in heat content:
Net change in heat content = heat added by hot water - heat removed by cold water
5. Set the net change in heat content equal to the initial heat content and solve for time:
Net change in heat content = initial heat content
time = initial heat content / (heat added by hot water - heat removed by cold water)
Plug in the values and calculate the time.
Please note that we need to convert the temperature to Celsius and the time to seconds to ensure consistency in units.
To solve this problem, we need to consider the principle of conservation of energy. The total energy of the water in the bath will remain constant throughout the process. We can determine the total energy using the specific heat capacity formula:
Q = mcΔT
where:
Q is the energy transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the temperature change
Given:
Initial water temperature (T1) = 60°C
Mass of water (m1) = 100 kg
Hot water flow rate (m2) = 20 kg/min
Hot water temperature (T2) = 70°C
Cold water flow rate (m3) = 20 kg/min
Cold water temperature (T3) = 10°C
Target temperature (T4) = 45°C
First, calculate the initial energy of the water in the bath:
E1 = m1c(T1 - T4)
Next, calculate the energy added by the hot water flowing into the bath over time:
E2 = m2c(T2 - T4)t
where t is the time in minutes. The energy added by the cold water flowing into the bath is:
E3 = m3c(T3 - T4)t
According to the principle of conservation of energy, the initial energy plus the energy added by the hot water plus the energy added by the cold water equals the final energy:
E1 + E2 + E3 = 0
Substituting the given values, we get:
100c(60 - 45) + 20c(70 - 45)t + 20c(10 - 45)t = 0
Simplifying the equation further:
1500c + 25(50c)t - 35(35c)t = 0
Dividing the equation by c(1500 + 25(50)t - 35(35)t) yields:
t = -2000/(-875t + 1500)
To find the time it will take for the water temperature to drop to 45°C, we can now solve this equation. By trial and error or using numerical methods, we can approximate the value of t.