Native Canadians used to boil water from maple tree sap by heating rocks and placing them into vats of sap sitting in hollowed-out tree logs. For this problem, assume that the syrup and the rocks have the heat capacities of water and sand respectively. What mass of stone, heated to 1000 °C, would be needed to increase 20.0 kg of sap from 60.0 °C to 85.0 °C? (Assume both the sap and stone come to the same final temperature)
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- (a) How much heat in joules is gained by the water?(b) How much heat in joules is lost by the metal?(c) What is the heat capacity of this piece of metal?A) If the final temperature is 22.1°C, what is the specific heat of the object? (2 sig fig) B) Referring to the table, identify the material in the objectThe same amount of heat entering identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.70 kcal of heat enters 1.38 kg of the following, originally at 30.2°C. The specific heat capacity for each material is given in square brackets below. (a) water [1.00 kcal/(kg · °C)] °C (b) concrete [0.20 kcal/(kg · °C)] °C (c) steel [0.108 kcal/(kg · °C)] °C (d) mercury [0.0333 kcal/(kg · °C)] °C
- A heavy pot made of copper has a mass of 2.07 kg (including the lid) and is heated in an oven to a temperature of 155 °C. You pour 0.10 kg of water at 25.6 °C into the pot and quickly close the lid so that no steam can escape. We assume that no heat is lost to the surrounding. For copper, Ccopper 390 J/(kg.K) = For water, Cwater = 4190 J/(kg.K), Lv = 2256 kJ/kg, Lƒ = 333 kJ/kg. What is the final mass of steam in the pot? gA 390-g metal container, insulated on the outside, holds 170.0 g of water in thermal equilibrium at 21.0°C. A 18.0-g ice cube, at -15.0°C, is dropped into the water, and when thermal equilibrium is reached the temperature is 12.0°C. Assume there is no heat exchange with the surroundings. The specific heat capacity of water is 4190 J/kg ∙ K, the specific heat capacity of ice is 2090 J/kg ∙ K and the heat of fusion is 3.34 × 105 J/kg. What is the specific heat capacity of the metal of the container?The same amount of heat entering identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.40 kcal of heat enters 1.83 kg of the following, originally at 28.2°C. The specific heat capacity for each material is given in square brackets below. (a) water [1.00 kcal/(kg · °C)] °C (b) concrete [0.20 kcal/(kg · °C)] °C (c) steel [0.108 kcal/(kg · °C)] °C (d) mercury [0.0333 kcal/(kg · °C)] °C
- A heavy pot made of copper has a mass of 2.24 kg (including the lid) and is heated in an oven to a temperature of 154 °C. You pour 0.12 kg of water at 26.0 °C into the pot and quickly close the lid so that no steam can escape. We assume that no heat is lost to the surrounding. For copper, Ccopper 390 J/(kg.K) For water, Cwater 4190 J/(kg.K), L, = 2256 kJ/kg, L; = 333 kJ/kg. %3D What is the final mass of steam in the pot? onA cube of ice is taken from the freezer at -9.5 °C and placed in a 95-g aluminum calorimeter filled with 300 g of water at room temperature of 20.0 °C. The final situation is observed to be all water at 16.0 °C. The specific heat of ice is 2100 J/kg · C°, the specific heat of aluminum is 900 J/kg · C°, the specific heat of water is is 4186 J/kg·C°, the heat of fusion of water is 333 kJ/Kg. Part A What was the mass of the ice cube? Express your answer to two significant figures and include the appropriate units. ? Value Units m = Submit Request AnswerIce at 0 °C is placed in a Styrofoam cup containing 0.62 kg of lemonade at 32 °C. The specific heat capacity of lemonade is virtually the same as that of water; that is, c = 4180 J/(kg C°). After the ice and lemonade reach an equilibrium temperature, some ice still remains. The latent heat of fusion for water is Lf = 3.35 x 105 J/ kg. Assume that the mass of the cup is so small that it absorbs a negligible amount of heat, and ignore any heat lost to the surroundings. Determine the mass of ice that has melted, in grams.