It actually depends on the temperature of the ice, but we’ll assume exactly 0C. You’ll need to use the heat of fusion of ice to determine how much energy it’ll take to melt it. Add that energy to the water (using the mc’delta’T equation) to determine how much that energy will lower the temperature of the water. Now you’ll have .5 kg of water at 0C being added to the 5L of water at whatever temperature it currently is (you’ll need to determine the mass of 5L of water). Again, use the calorimetry equation to combine them, and you’ll have your final answer.

that’s greater of a thermal physics than a chemistry concern. The water could desire to be heated via seventy 5 C to boil (we predict sea point). you are able to calculate the completed potential required to try this from the quantity of water and its warmth potential. Dividing this via 750 watts provide you with the time. The photon portion of the situation demands some common quantum mechanics, particularly the formula E = h / lambda, the place E is the potential of a photon, h is Planck’s consistent, and lambda is the wavelength of the radiation (given — incorrectly — that’s going to be 14.3 cm.). seem up Planck’s consistent, do the obtrusive branch, and get the potential in line with photon. you comprehend the completed potential, so one branch provide you with the style of photons required to hold that lots potential. that’s a sizeable quantity.