To determine the rate of change of entropy with respect to time, we can use the equation:dS/dt = dQ/dt * 1/T where dS/dt is the rate of change of entropy, dQ/dt is the rate of heat transfer, and T is the temperature in Kelvin.First, we need to find the rate of heat transfer dQ/dt . We can use the equation:dQ/dt = mc * dT/dt where m is the mass of the water, c is the specific heat capacity of water, and dT/dt is the rate of temperature change.Given that the water is heated at a constant rate of 2C/min, we can convert this to Kelvin per second:dT/dt = 2 K/min * 1 min/60 s = 1/30 K/sThe specific heat capacity of water is approximately 4.18 J/ g*K . Assuming we have 100 g of water in the beaker, we can calculate dQ/dt:dQ/dt = 100 g * 4.18 J/ g*K * 1/30 K/s = 13.93 J/sNow we can find the rate of entropy production as the water is heated from 25C to 100C. Since the temperature is changing, we need to integrate the equation for dS/dt over the temperature range:S = dQ/dt * 1/T dTWe can substitute the values we found for dQ/dt and the temperature range in Kelvin 298 K to 373 K :S = 13.93 J/s * 1/T dT from 298 K to 373 KTo solve this integral, we can use the natural logarithm function:S = 13.93 J/s * ln 373 K - ln 298 K = 13.93 J/s * ln 373/298 S 13.93 J/s * 0.223 = 3.11 J/ K*s So, the rate of entropy production in the water as it is heated from 25C to 100C at a constant rate of 2C/min is approximately 3.11 J/ K*s .