Question
Coffee in a closed plastic foam cup is sitting on a table. The inside of the cup has a temperature of 60ºC. Once sugar and milk are added to the cup, the temperature inside the cup changes to 49ºC.
Equation 1 shows a mathematical model for the energy transferred.
Equation 1. Energy Transfer in a System
Q = m c Δ T
Q is the amount of energy transferred, in Joules (J)
m is the mass, in grams (g)
c is the specific heat of coffee, which equals 4.184 in Joules per gram degree Celsius (J/gºC)
ΔT
is the change in temperature, in degrees Celsius (ºC)
Table 1. Mass of Contents Inside of the Plastic Foam Cup
Coffee Sugar Milk
114 g 4 g 37 g
Calculate the amount of energy, in Joules (J), that is transferred inside the plastic foam cup. Choose the whole number value from the first drop down menu. Then, click on the blank boxes and select words or phrases to identify the direction of energy transfer inside the cup.
(2 points)
The number of Joules transferred is
. The energy was transferred from the
to the milk and sugar.
Question 2
Part B
Match the independent variables, dependent variable, and control for an experiment that could determine the heat transfer in a different cup of coffee with milk and sugar added.
(5 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
Specific heat capacity of coffee (c)
Mass of coffee (g)
Mass of sugar (g)
Mass of milk (g)
Temperature change inside the cup (ΔT
)
Equation 1 shows a mathematical model for the energy transferred.
Equation 1. Energy Transfer in a System
Q = m c Δ T
Q is the amount of energy transferred, in Joules (J)
m is the mass, in grams (g)
c is the specific heat of coffee, which equals 4.184 in Joules per gram degree Celsius (J/gºC)
ΔT
is the change in temperature, in degrees Celsius (ºC)
Table 1. Mass of Contents Inside of the Plastic Foam Cup
Coffee Sugar Milk
114 g 4 g 37 g
Calculate the amount of energy, in Joules (J), that is transferred inside the plastic foam cup. Choose the whole number value from the first drop down menu. Then, click on the blank boxes and select words or phrases to identify the direction of energy transfer inside the cup.
(2 points)
The number of Joules transferred is
. The energy was transferred from the
to the milk and sugar.
Question 2
Part B
Match the independent variables, dependent variable, and control for an experiment that could determine the heat transfer in a different cup of coffee with milk and sugar added.
(5 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
Specific heat capacity of coffee (c)
Mass of coffee (g)
Mass of sugar (g)
Mass of milk (g)
Temperature change inside the cup (ΔT
)
Answers
GPT-4o mini
To calculate the amount of energy transferred in the plastic foam cup when the coffee is mixed with sugar and milk, we need to use the equation:
\[ Q = m c \Delta T \]
### Given Data:
- Mass of coffee (m) = 114 g
- Mass of sugar (m) = 4 g
- Mass of milk (m) = 37 g
- Total mass (m_total) = 114 g + 4 g + 37 g = 155 g
- Initial temperature of coffee (T_initial) = 60ºC
- Final temperature after adding sugar and milk (T_final) = 49ºC
- Specific heat of coffee (c) = 4.184 J/gºC
### Change in Temperature (ΔT):
\[ \Delta T = T_{final} - T_{initial} = 49ºC - 60ºC = -11ºC \]
### Energy Transferred Calculation:
Using the formula:
\[ Q = 155 \, \text{g} \times 4.184 \, \text{J/gºC} \times (-11ºC) \]
\[ Q = 155 \times 4.184 \times -11 \]
Calculating:
1. First, calculate \( 155 \times 4.184 \):
\[ 155 \times 4.184 = 648.92 \, \text{J/ºC} \]
2. Now, multiply that by -11:
\[ 648.92 \times -11 = -7138.12 \, \text{J} \]
### Final Result:
Since energy transferred can be considered as a positive amount but indicates heat lost by the coffee:
The number of Joules transferred is **7138** (rounded to the whole number). The energy was transferred from the **coffee** to the milk and sugar.
---
### Question 2: Experiment Variables
- **Independent Variables:**
- Mass of coffee (g)
- Mass of sugar (g)
- Mass of milk (g)
- **Dependent Variable:**
- Temperature change inside the cup (ΔT)
- **Control:**
- Specific heat capacity of coffee (c)
You can construct your input as follows:
- **Independent Variables:**
- Mass of coffee (g)
- Mass of sugar (g)
- Mass of milk (g)
- **Dependent Variable:**
- Temperature change inside the cup (ΔT)
- **Control:**
- Specific heat capacity of coffee (c)
This configuration helps isolate the effect of the independent variables on the dependent variable while keeping the controlled variable constant.
\[ Q = m c \Delta T \]
### Given Data:
- Mass of coffee (m) = 114 g
- Mass of sugar (m) = 4 g
- Mass of milk (m) = 37 g
- Total mass (m_total) = 114 g + 4 g + 37 g = 155 g
- Initial temperature of coffee (T_initial) = 60ºC
- Final temperature after adding sugar and milk (T_final) = 49ºC
- Specific heat of coffee (c) = 4.184 J/gºC
### Change in Temperature (ΔT):
\[ \Delta T = T_{final} - T_{initial} = 49ºC - 60ºC = -11ºC \]
### Energy Transferred Calculation:
Using the formula:
\[ Q = 155 \, \text{g} \times 4.184 \, \text{J/gºC} \times (-11ºC) \]
\[ Q = 155 \times 4.184 \times -11 \]
Calculating:
1. First, calculate \( 155 \times 4.184 \):
\[ 155 \times 4.184 = 648.92 \, \text{J/ºC} \]
2. Now, multiply that by -11:
\[ 648.92 \times -11 = -7138.12 \, \text{J} \]
### Final Result:
Since energy transferred can be considered as a positive amount but indicates heat lost by the coffee:
The number of Joules transferred is **7138** (rounded to the whole number). The energy was transferred from the **coffee** to the milk and sugar.
---
### Question 2: Experiment Variables
- **Independent Variables:**
- Mass of coffee (g)
- Mass of sugar (g)
- Mass of milk (g)
- **Dependent Variable:**
- Temperature change inside the cup (ΔT)
- **Control:**
- Specific heat capacity of coffee (c)
You can construct your input as follows:
- **Independent Variables:**
- Mass of coffee (g)
- Mass of sugar (g)
- Mass of milk (g)
- **Dependent Variable:**
- Temperature change inside the cup (ΔT)
- **Control:**
- Specific heat capacity of coffee (c)
This configuration helps isolate the effect of the independent variables on the dependent variable while keeping the controlled variable constant.