Initial temperature hot tea 52° C,
Final temperature after adding milk 39° C.
Initial temperature of cold milk is 19° C
Quantity hot tea 175 ml
c) What is your calculated heat transfer (Q) of the hot beverage when 4 ounces (120 mL) of cold beverage was added? Report your answer in Joules.
My answer: 0.12ml * (4.18 kj/kg.C) * (39C - 19C)= 10.032 J
d)What is your calculated heat transfer (Q) of the cold beverage when 4 ounces (120 mL) was added to the hot beverage? Report your answer in Joules.
My Answer 0.175ml * (3.85 kj/kg.C) * (39C - 52C)= −8.75875 J
correct except for units. I am not keen on your use of units, wthout some assumption of density (ie liters to kg)
To calculate the heat transfer (Q) of the hot beverage when 4 ounces (120 mL) of cold beverage was added, you can use the formula:
Q = m * c * ΔT
Where:
m is the mass of the cold beverage (in kg), which can be calculated by converting the volume (in mL) to kg using the density of the cold beverage (assumed to be the same as water, which is 1 g/mL or 1 kg/L).
c is the specific heat capacity of the hot beverage (in J/kg·°C), which represents how much heat is needed to raise the temperature of 1 kg of the substance by 1 degree Celsius.
ΔT is the change in temperature (in °C) of the hot beverage.
Given:
m (mass of the cold beverage) = 0.120 kg (converted from 120 mL)
c (specific heat capacity of the hot beverage) = 4.18 kJ/kg·°C = 4.18 × 10^3 J/kg·°C (converted from kJ to J)
ΔT (change in temperature of the hot beverage) = 39°C - 19°C = 20°C
Now, substitute the values into the formula:
Q = 0.120 kg * (4.18 × 10^3 J/kg·°C) * 20°C
Q = 10.032 J
Therefore, the calculated heat transfer (Q) of the hot beverage when 4 ounces (120 mL) of cold beverage was added is 10.032 J.
For the calculated heat transfer (Q) of the cold beverage when 4 ounces (120 mL) was added to the hot beverage, the same formula can be used. However, the specific heat capacity (c) and change in temperature (ΔT) will be different for the cold beverage.
Given:
m (mass of the hot beverage) = 0.175 kg (converted from 175 mL)
c (specific heat capacity of the cold beverage) = 3.85 kJ/kg·°C = 3.85 × 10^3 J/kg·°C (converted from kJ to J)
ΔT (change in temperature of the cold beverage) = 39°C - 52°C = -13°C
Now, substitute the values into the formula:
Q = 0.175 kg * (3.85 × 10^3 J/kg·°C) * (-13°C)
Q = -8.75875 J
Therefore, the calculated heat transfer (Q) of the cold beverage when 4 ounces (120 mL) was added to the hot beverage is -8.75875 J.