Final answer:
It will take approximately 159 seconds to heat 250 g (250 mL) of water from 18°C to 75°C with a 375-watt immersion heater, given the specific heat of water is 4.184 J/g°C.
Step-by-step explanation:
The student asked how long it will take to heat a cup of soup, assumed to be 250 mL of water, from 18°C to 75°C using a small immersion heater rated at 375 watts. To answer this, we use the concept of specific heat capacity and energy transfer. The specific heat of water is 4.184 J/g°C. The energy required (Q) can be calculated with the formula Q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
First, convert the volume of water to mass: 250 mL of water is equivalent to 250 g. The change in temperature (ΔT) is 75°C - 18°C = 57°C. The energy required is therefore 250 g * 4.184 J/g°C * 57°C = 59,589 J. Given the heater's power is 375 W (375 J/s), the time (t) it would take is given by t = Q / Power, so t = 59,589 J / 375 J/s which results in approximately 159 seconds.