7150/(33.5e4) = g ice doesn't work.
Am I calculating incorrectly?
heat of fusion for water is 334 J/g. You want it in J/g since q is given in the problem in J. The answer will be in units of grams of ice. You will need to change to kg since the problem asks for it in kg.
Should I get 0.0214 kilograms? Because this answer is listed as incorrect.
q=heat of fusion x g ice
You now q. You know heat of fusion (or can look it up). Solve for g ice.
A Carnot engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. When 7150 J of heat is put into the engine and the engine produces work, how many kilograms of ice in the tub are melted due to the heat delivered to the cold reservoir?
I get the same answer you have. Bob Pursley is on-line but away from his computer at the moment. I will email this to him and let him take a look at it. I don't know exactly when he will be back.
First determine the efficiency:
efficiency= (373-273)/373= .268
That means that 1-.268 is waste heat
Now if 7150J is put into the engine, some is delivered as work, and the rest is waste delivered to the ice.
MassIce melted= (1-.268)7150J/(334J/Kg)
To calculate the mass of ice melted due to the heat delivered to the cold reservoir, you can follow these steps:
1. Determine the efficiency of the Carnot engine. Efficiency is defined as the ratio of the work output to the heat input. In this case, the efficiency is given by:
efficiency = (T_hot - T_cold) / T_hot
With T_hot being the temperature of the hot reservoir (in Kelvin) and T_cold being the temperature of the cold reservoir (in Kelvin). Since boiling water is used as the hot reservoir, T_hot = 373 K and since it is a large tub of ice and water, T_cold = 273 K.
efficiency = (373 - 273) / 373 = 0.268 (or 26.8%)
2. Calculate the amount of waste heat delivered to the ice. Since the total heat input is 7150 J, the work done by the engine is given by:
work = efficiency * heat input = 0.268 * 7150 J = 1914.2 J
The remaining heat (waste heat) delivered to the cold reservoir (ice) is the difference between the total heat input and the work done:
waste heat = heat input - work = 7150 J - 1914.2 J = 5235.8 J
3. Convert the waste heat to the mass of ice melted. Use the heat of fusion for water, which is given as 334 J/g (joules per gram). Since the question asks for the answer in kilograms, convert the waste heat to grams by dividing by 1000:
mass of ice melted = waste heat / heat of fusion = 5235.8 J / 334 J/g = 15.66 g
Finally, convert the mass from grams to kilograms by dividing by 1000:
mass of ice melted = 15.66 g / 1000 = 0.01566 kg
So, the correct answer is approximately 0.01566 kilograms of ice melted due to the heat delivered to the cold reservoir.