To find out after how many minutes the coffee will be lukewarm (30°C), we can set up the equation based on the given temperature model:

$H = 93(0.97)^t$

We need to solve for $t$ when $H = 30$:

$30 = 93(0.97)^t$

Now, divide both sides by 93:

$\frac{30}{93} = (0.97)^t$

Calculating $\frac{30}{93}$:

$\frac{30}{93} \approx 0.32258$

Now we have:

$0.32258 = (0.97)^t$

To solve for $t$, we take the logarithm of both sides:

$\log(0.32258) = \log((0.97)^t)$

Using the power rule of logarithms:

$\log(0.32258) = t \cdot \log(0.97)$

Now, solve for $t$:

$t = \frac{\log(0.32258)}{\log(0.97)}$

Calculating the logarithms:

1. $\log(0.32258) \approx -0.491$
2. $\log(0.97) \approx -0.013$

Now substitute these values:

$t \approx \frac{-0.491}{-0.013} \approx 37.77$

Therefore, the coffee will be lukewarm after approximately 37.77 minutes.

Summary

• $t \approx 37.77$ minutes

Would you like any more details on the calculations or the model? Here are some related questions you might consider:

1. How does the cooling rate change with different initial temperatures?
2. What would happen to the cooling rate on a warmer day?
3. How can we model other cooling scenarios using similar equations?
4. What assumptions are made in the cooling model?
5. How does the environment (like wind or sun) affect the cooling process?

Tip: When working with logarithmic equations, always check if the base and values are in the appropriate range to ensure valid results.