8.11 : How much energy will be required if a mass of 100 kg escapes from the earth ? [R.= 6.4 x 10° m, g 10 m/s²] %3D (a) 3.2 x 10° J (b) 6.4 x 10°J (c) 1.6 x 10° J (d) 8 x 10° I
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- You recently bought an aluminum (C = 0.9 J/g-K) kettle with a mass of 2 kg. You used the kettle to heat water (C = 4.184 J/g-K). The kettle contains water (1.5 kg). The kettle was heated using an induction stove from 20°C to 85°C. The water is then used for coffee. If no heat is lost to the surroundings, what is the total heat added to raise the temperature?A black wood stove has a surface area of 2.00 m2 and a surface temperature of 146°C. What is the net rate at which heat is radiated into the room? The room temperature is 20.0°C. Stefan–Boltzmann constant is 5.670 × 10−8 W/(m2·K4). in kWConsider 1 g of sodium chloride that is successively divided into smaller cubes with various varying sizes as shown in table. Assuming a mass density of 2.178×10³ kg/m³ and surface energy of 2 × 10-5 J/cm², complete the table below and make your conclusion about how the total surface energy of 1 g of sodium chloride vary with particle size. Total surface area Side 0.1 cm 0.01 cm 0.001 cm 1 μm 1 nm Total Surface energy (J/g)
- A block of ice (m = 5 kg) at a temperature of T1 = 0 degrees C is placed out in the sun until it melts, and the temperature of the resulting water rises to T2 = 28 degrees C. Recall that the specific heat of water is c = 4186 J/(kg⋅K), and its latent heat of fusion is Lf = 3.34 × 105 J/kg. a.) Input an expression for the amount of energy, Em, needed to melt the ice into water. b.) Input an expression for the total amount of energy, Etot, to melt the ice and then bring the water to T2. c.) What is this energy in joules?1.5 kg of water (c = 4189 J/(kg⋅K)) is heated from T1 = 11° C to T2 = 20.5° C. IF 0.5 kg of water at T3 = 20° C is added to the heated water. What is the final temperature of the water, in units of kelvin?How much energy does a freezer have to remove from 1.5-kg of water at 20^0C to turn it into ice at -12^0C? specific heat capacity of water is 4184 \frac{J}{kg^0C) specific heat capacity of ice = 2090 \frac{J}{kg^0C) heat of fusion of ice = 3.33x10^5\ \frac{J}{kg}?
- The energy reaching Earth from the Sun at the top of the atmosphere is 1.36 x10^3 W/m2, called the solar constant. Assuming that Earth radiates like a blackbody at uniform temperature, what do you conclude is the equilibrium temperature of Earth?Large meteors sometimes strike the Earth, converting most of their kinetic energy into thermal energy. If a 10° kg meteor moving at 25.0 km/s lands in a deep ocean and 80% of its kinetic energy goes into heating water, how many kilograms of water could it raise by 5.0°C? Select the correct answer O 8 x 1012 kg O 5 x 1013 kg O 1 x 1013 kg O 4 x 1012 kg. Your Answer O 2 x 1013 kgA 240 m/s projectile with specific heat c = 0.128 J/gK lodges in a slab of ice at 00C. How much of the ice melts if the projectile is initially at 350C and has a mass of 4.5 g?
- An open container holds 0.550 kg of ice at -15.0∘C. The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 800.0 J/min for 500.0 min. After how many minutes does the ice start to melt? Express your answer in minutes. After how many minutes, from the time when the heating is first started, does the temperature begin to rise above 0.0∘C?A cup of hot coffee sits on your desk. Assume that the thermal energy in the coffee can be converted with an efficiency of 100%% to gravitational potential energy of the coffee-Earth system-enough so that the coffee rises up above the rim of the cup. Let the height of the coffee rise by 3.4 cm. Find the magnitude of the drop in temperature needed to accomplish this energy conversion. Use c = 4180 J/kg⋅C for coffee.After sitting on a shelf for a while, a can of soda at a room temperature (73∘73∘F) is placed inside a refrigerator and slowly cools. The temperature of the refrigerator is 37∘37∘F. Newton's Law of Cooling explains that the temperature of the can of soda will decrease proportionally to the difference between the temperature of the can of soda and the temperature of the refrigerator, as given by the formula below: �=��+(�0−��)�−��T=Ta+(T0−Ta)e−kt ��=Ta= the temperature surrounding the object�0=T0= the initial temperature of the object�=t= the time in minutes�=T= the temperature of the object after �t minutes�=k= decay constant The can of soda reaches the temperature of 53∘53∘F after 30 minutes. Using this information, find the value of �k, to the nearest thousandth. Use the resulting equation to determine the Fahrenheit temperature of the can of soda, to the nearest degree, after 60 minutes. Enter only the final temperature into the input box.