You place frozen soup (T = −17°C) in a microwave oven for 3.5 min. The oven transfers 650 W to the soup. Model the soup as 0.35 kg of water (initially ice). What are the temperature and state of the soup when the oven stops?
The correct answer and explanation is :
To determine the temperature and state of the soup after being heated in the microwave for 3.5 minutes, we’ll need to calculate the energy transferred, how that energy is used to heat the soup, and whether it results in the soup melting or not. The process consists of two phases: heating the ice to 0°C and then melting the ice into water.
Given:
- Initial temperature of soup (T₁) = −17°C
- Mass of soup (m) = 0.35 kg
- Power transferred (P) = 650 W
- Time (t) = 3.5 min = 210 seconds
- Latent heat of fusion of ice (L_f) = 334,000 J/kg
- Specific heat of water (cₓ) = 4,186 J/(kg·°C)
- Specific heat of ice (c_ice) = 2,090 J/(kg·°C)
Step 1: Calculate the total energy transferred
The energy transferred to the soup is the microwave power multiplied by the time:
[
Q = P \times t = 650 \, \text{W} \times 210 \, \text{s} = 136,500 \, \text{J}
]
Step 2: Heat the soup to 0°C
First, the ice at −17°C must be warmed to 0°C. The energy required for this is calculated using the formula:
[
Q_1 = m \times c_{\text{ice}} \times \Delta T = 0.35 \, \text{kg} \times 2,090 \, \text{J/kg°C} \times (0 – (-17)) \, \text{°C}
]
[
Q_1 = 0.35 \times 2,090 \times 17 = 12,402 \, \text{J}
]
So, 12,402 J is used to raise the temperature of the ice from −17°C to 0°C.
Step 3: Melting the ice
After reaching 0°C, the ice starts melting. The energy required to melt the ice is:
[
Q_2 = m \times L_f = 0.35 \, \text{kg} \times 334,000 \, \text{J/kg} = 116,900 \, \text{J}
]
Step 4: Determine if the ice melts
The total energy available is 136,500 J. We first use 12,402 J to heat the ice to 0°C, leaving:
[
136,500 – 12,402 = 124,098 \, \text{J}
]
Since 116,900 J is needed to melt the ice, and we have 124,098 J available, all of the ice will melt. The remaining energy will be:
[
124,098 – 116,900 = 7,198 \, \text{J}
]
Step 5: Final temperature of the soup
After the ice melts, we have 0.35 kg of water at 0°C, and there is still 7,198 J of energy available to heat the water. The temperature change in the water is given by:
[
Q_3 = m \times c_{\text{water}} \times \Delta T
]
[
7,198 \, \text{J} = 0.35 \, \text{kg} \times 4,186 \, \text{J/kg°C} \times \Delta T
]
Solving for (\Delta T):
[
\Delta T = \frac{7,198}{0.35 \times 4,186} \approx 4.88 \, \text{°C}
]
Therefore, the final temperature of the water will be approximately:
[
T_{\text{final}} = 0°C + 4.88°C \approx 4.88°C
]
Conclusion:
After 3.5 minutes of heating, the soup will be in a liquid state (water) at approximately 4.88°C.