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A small immersion heater is rated at 255 W . The specific heat of water is 4186 J/kg⋅C°.
A small immersion heater is rated at 255 W . The specific heat of water is 4186 J/kg⋅C°.Estimate how long it will take to heat a cup of soup (assume this is 250 ml of water) from 17 °C to 62 °C. Ignore the heat loss to the surrounding environment.
Final answer:The calculation requires thephysics formula:Q = mcΔT. The weight of the water is 0.25 kg and temperature difference is 45°C. Using these values and the power of the heater, the soup will take approximately 185 seconds, or just over 3 minutes, to heat up.Explanation:The calculation you're dealing with here translates to aphysicsproblem that requires the use of the formula:Q = mcΔT. Firstly, convert 250 ml to kilograms, given that the density of water is approximately 1 kg/L. So, 250 ml = 0.25 kg. Subsequentlycalculate the temperature difference, which is ΔT = 62°C - 17°C = 45°C. Apply the values to the formula: Q = 0.25 Kg * 4186 J/kg°C * 45°C = 47107.5 J.Thepower of the heateris 255W, which is 255 J/s. Thus, time can be calculated as Q/Power = 47107.5 J / 255 J/S = ~185 seconds or approximately 3 minutes and 5 seconds.Learn more about Heat Measurement here:brainly.com/question/18912282#SPJ12...