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This problem involves Newton's law of cooling, which describes how an object's temperature changes over time. The formula given is:

\[
T(t) = T_a + (T_0 - T_a)e^{-kt}
\]

Where:
- \( T(t) \) is the temperature of the object after time \( t \).
- \( T_a \) is the ambient (surrounding) temperature, which in this case is 80°F.
- \( T_0 \) is the initial temperature of the object, which is 131°F.
- \( k \) is the cooling constant, given as 0.00348.
- \( t \) is the time in hours (24 hours in this case).

### Given:
- \( T_0 = 131^\circ F \)
- \( T_a = 80^\circ F \)
- \( k = 0.00348 \)
- \( t = 24 \) hours

We need to calculate the temperature of the water after 24 hours, \( T(24) \), to check if it's still warm enough for Dominic's shower (at least 116°F).

Let's calculate this!The temperature of the water after 24 hours will be approximately 126.91°F. Since Dominic prefers the water to be at least 116°F for a shower, the water will still be warm enough.

The correct answer is \( \text{b. Yes, the temperature after 24 hours will be approximately 127°F} \).

Would you like more details on the calculations or have any questions?

Here are some related questions:
1. How does changing the cooling constant \( k \) affect the cooling rate?
2. What would the temperature be after 12 hours?
3. What happens if the ambient temperature \( T_a \) increases or decreases?
4. How long would it take for the water to cool to 100°F?
5. How does insulation impact the value of \( k \) in real-world scenarios?

**Tip:** The cooling constant \( k \) reflects how well an object retains or loses heat. Higher insulation leads to a smaller \( k \), meaning slower heat loss.

Newton's law of cooling indicates that the temperature of a warm object will decrease exponentially with time and will approach the temperature of the surrounding air. The temperature T(t) is modeled by T(t) = Ta + (T0 - Ta)e^-kt. In this model, Ta represents the temperature of the surrounding air, T0 represents the initial temperature of the object, and t is the time after the object starts cooling. The value of k is the cooling rate and is a constant related to the physical properties of the object. Water in a water heater is originally 131°F. The water heater is shut off and the water cools to the temperature of the surrounding air, which is 80°F. The water cools slowly because of the insulation inside the heater, and the rate of cooling is 0.00348. Dominic does not like to shower with water less than 116°F. If Dominic waits 24 hr after the heater is shut off, will the water still be warm enough for a shower? How warm will the water be?

Learn how to apply Newton's Law of Cooling to calculate the temperature of water after 24 hours in a heater that is initially 131°F, with a cooling constant of 0.00348. Understand the exponential decay process and whether the water will remain warm enough for a shower.

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