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In different experiment, a liquid must be cooled 6 times as fast as the liquid in the example, but it must still start at 0C. What will the change in temperature be each hour?

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To cool the liquid 6 times faster, the change in temperature each hour will be 1/6th of the change in temperature in the example.

To solve this problem, we can use the formula: Q = mcΔT, where Q is the amount of heat, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

Since the liquid must be cooled 6 times faster, it means that the change in temperature each hour will be 1/6th of the change in temperature in the example.

Let's say the change in temperature in the example is ΔT1. The change in temperature each hour in the new experiment would be

Therefore, the change in temperature each hour in the new experiment would be .

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