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How long does it take a 19 kW steam engine to do \( 6.8 \times 10^{7} \mathrm{~J} \) of work? \( (\mathrm{k}=\mathrm{kilo}=1000) \)

Ask by Wilson Lindsey. in the United States
Mar 12,2025

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The engine takes approximately 3579 seconds or about 59.65 minutes to do \(6.8 \times 10^{7} \, \text{J}\) of work.

Solución

We are given the work \[ W = 6.8 \times 10^{7} \, \text{J} \] and the power of the engine \[ P = 19 \, \text{kW} = 19 \times 1000 \, \text{W} = 19000 \, \text{W}. \] Since power is defined as work over time, \[ P = \frac{W}{t}, \] we can solve for time \( t \) as follows: \[ t = \frac{W}{P} = \frac{6.8 \times 10^{7} \, \text{J}}{19000 \, \text{W}}. \] Now, perform the division: \[ t = \frac{6.8 \times 10^{7}}{19000} \approx 3578.95 \, \text{seconds}. \] To convert seconds to minutes, note that 1 minute equals 60 seconds: \[ t \approx \frac{3578.95}{60} \approx 59.65 \, \text{minutes}. \] Thus, the engine takes approximately 3579 seconds or about 59.65 minutes to do \(6.8 \times 10^{7} \, \text{J}\) of work.

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To find out how long it takes for a 19 kW steam engine to do \( 6.8 \times 10^{7} \) J of work, we can use the formula: \[ \text{Power (P)} = \frac{\text{Work (W)}}{\text{Time (t)}} \] Rearranging gives us: \[ t = \frac{W}{P} \] Substituting the values: \[ t = \frac{6.8 \times 10^{7} \text{ J}}{19 \times 10^{3} \text{ W}} \\ t \approx \frac{6.8 \times 10^{7}}{19000} \\ t \approx 3587.37 \text{ seconds} \] So, it would take approximately 3587 seconds, or about 59.8 minutes, for the engine to perform that amount of work. For historical enthusiasts, the steam engine revolutionized industries in the 18th and 19th centuries, fueling not just factories but also railways and ships, ushering in the Industrial Revolution—an era that reshaped economies and social structures globally! If you're ever in a jam while tackling such problems, remember to double-check your unit conversions, as a small mistake can lead to big discrepancies in your answer. Always ensure you're working with consistent units—like Watts with Joules and seconds—to keep your calculations accurate and flowing smoothly!

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