A flow calorimeter is used to measure the specific heat of a liquid. Heat is added at a known rate to a stream of the liquid as it passes through the calorimeter at a known rate. Then a measurement of the resulting temperature difference between the inflow and the outflow points of the liquid stream enables one to compute the specific heat of the liquid. A liquid of density 0.85 g/cm^3 flows through a calorimeter at a rate of 7.80 cm^3/s. Heat is added by means of a 266 watt heater coil, and a temperature difference of 11.0 degrees C is established in steady-state conditions between the inflow and outflow points. Find the specific heat of the liquid (in joules/kilogram-degree C).
To find the specific heat of the liquid in joules/kilogram-degree Celsius, we can follow these steps:
Step 1: Calculate the mass flow rate of the liquid.
Density = 0.85 g/cm³
Flow rate = 7.80 cm³/s
To convert cm³ to kg, we need to divide by 1000 (since 1 cm³ = 0.001 kg).
Mass flow rate = (Density * Flow rate) / 1000 = (0.85 g/cm³ * 7.80 cm³/s) / 1000 = 0.00663 kg/s
Step 2: Calculate the power.
Power = 266 W
Step 3: Calculate the heat transferred per second.
Heat transferred per second = Power = 266 J/s
Step 4: Calculate the temperature difference in Kelvin.
Temperature difference = 11.0 °C
To convert Celsius to Kelvin, add 273.15.
Temperature difference in Kelvin = 11.0 °C + 273.15 = 284.15 K
Step 5: Calculate the specific heat.
Specific heat = Heat transferred per second / (Temperature difference * Mass flow rate)
Specific heat = 266 J/s / (284.15 K * 0.00663 kg/s)
Calculating the specific heat:
Specific heat = 15.82 J/(kg·°C)
Therefore, the specific heat of the liquid is approximately 15.82 joules/kilogram-degree Celsius.
To find the specific heat of the liquid, we need to use the formula:
Specific heat (C) = heat added (Q) / (mass (m) * temperature difference (ΔT))
First, let's find the mass of the liquid flowing through the calorimeter per unit time:
Mass flow rate = density * volume flow rate
The volume flow rate is given as 7.80 cm^3/s. Since the density is given as 0.85 g/cm^3, we can convert the volume flow rate to a mass flow rate:
Mass flow rate = (0.85 g/cm^3) * (7.80 cm^3/s)
Now, let's convert the mass flow rate to kilograms:
Mass flow rate = (0.85 g/cm^3) * (7.80 cm^3/s) * (1 kg / 1000 g)
Mass flow rate = 0.00663 kg/s
Now we can calculate the heat added (Q) using the formula:
Heat added (Q) = power (P) * time (t)
The power is given as 266 watts. Since we want to find the specific heat in joules/kilogram-degree C, we need to use seconds as the unit of time. Let's assume we collect data for 60 seconds:
Heat added (Q) = (266 watts) * (60 seconds)
Heat added (Q) = 15,960 joules
Finally, we can substitute the values into the formula for specific heat:
Specific heat (C) = (15,960 joules) / (0.00663 kg/s * 11.0 degrees C)
Calculating this result:
Specific heat (C) ≈ 2,235 joules/kilogram-degree C
Therefore, the specific heat of the liquid is approximately 2,235 joules/kilogram-degree C.