Calculate the specific heat of the metal?
ΔT (for the water) = 46.3 °C - 23.8 °C = 22.5 °C. Heat lost by metal = heat gained by water = (25.0 g)(22.5 °C)(4.184 J/(g·°C)) = 2353.5 J. ΔT (for the metal) = 78.3 °C - 46.3 °C = 32.0 °C. Specific heat = (heat change)/((mass)(temperature change)) = (2353.5 J)/((5.00 g)(32.0 °C)) = 14.7 J/(g·°C).
How much heat (in joules) is required to raise the temperature of 30.0 kg of water from 18°C to 81°C?
Q(heat energy) = m*c* delta T where c is the specific heat capacity of ur material, in this case, water. c(water) = 4.18J/ g* K So Q = 30000g * 4.18 J/g*K *63K = 7900200J = 7900.2kJ
If a piece of ice at 0°C is put into a bucket of hot water at 100°C, then what will be the final temperature of the mixture (no mass is given)?
Enthalpy of fusion of water is 79.8 cal/g, and 1 cal by definition will raise 1 g of water by 1 °C.If you start with equal masses, melting 1 gram will consume 79.8 cal, cooling the gram of hot water to 10 °C will provide 90 cal, and warming the gram of just-melted water to 10 °C will consume 10 cal, for a total balance of 0.2 cal left, which is enough to warm 2g of water from 10 °C to 10.1 °C.
10 g of ice at -10 degrees Celsius is added to 10 g of water at 20 degrees Celsius. What is its mixing temperature?
The temperature of ice being less than 0°C adds a correction to the question for temperature of ice to be -10°C rather being 10°C.Taking the specific heat of ice as 0.5 cal/gm°C,latent heat as 80cal/gm and specific heat of water as 1cal/gm°C.Assuming water at0°C as reference heat calculation, the heat content of ice at -10°C=>the quantity of heat content=(-)10×0.5×10–10×80 cal=-850cal.That of 10gm of water at 20°C would be =10×1×20=200cal.So mixing the two would bring the heat content of total system as -850+200=-650cal.Finally the total heat energy of the system brings-650 cal below the 20 gm of water ice mixture.=> 650/80=8.125 gm of ice floats on 11.875gm of water at 0°C.Ans..
A 300 g block of copper at a temperature of 75°C is dropped into 650 g of water at 38°C. The water is containe?
Let final temperature be T°C. We have 0.300*386*(75-T) = {0.650*4186 + 0.200*840}*(T-38) or dividing by 0.3 we get 386*(75-T) = {0.650*13953 + 0.200*2800}(T-38) or - 386T + 28950 = 9629*(T- 38) or (9629+386)T = 394869 or T = 39.43 or 39.4 degree
Thermal equilibrium, PHYSICS HW?
1.5 C / minute = 1.5/60 = 0.025 C/s specific heats Al = 900 J/kgC H2O = 4186 J/kgC Since P = Q/T then... P = mc(rate of change) + mc(rate of change) Change g to kg... P = 0.3(900)(0.025) + 0.8(4186)(0.025) P = 90.5 W Hope that helps...
A 200g aluminum cup contains 800g of water in thermal equilibrium at 80 degrees C. The combination of the cup?
Heat lost by cup to drop 1 Deg. C= (mass1*sph1)=0.2kg X 900 J/kg =180J/min Heat lost by water to drop 1 Deg. C=(mass2*sph2)=0.8 X 4186 J/kg =3348 J/min Total heat loss per minute = 180+3348=3528 J/min Heat lost to decrease 1.5 degrees C= 3528 X 1.5=5292 J/min 1 Watt= 1J/sec Total watts= 5292/60 =88.2 W
Why does the dissolution of NH4NO3 in water result in a decrease in the temperature of the solution?
Solid ammonium nitrate is made up of ammonium (NH4+) and nitrate ions (NO3-) held together by ionic bonds in a closely-packed crystal lattice.When solid ammonium nitrate comes in contact with water, the polar water molecules will interact with these ions and attract individual ions from the lattice structure, which eventually breaks down. e.g;-NH4NO3(s) ——→ NH4+(aq) + NO3-(aq)To break the ionic bonds holding the lattice together requires energy, which is absorbed from the surrounding environment, making the solution cold.Some heat is generated when the ammonium and nitrate ions interact with the water molecules (exothermic reaction), but this heat is far less than that required for the water molecules to break the strong ionic bonds in the solid ammonium nitrate. Therefore the overall dissolution of ammonium nitrate in water is strongly endothermic.For this reason solid ammonium nitrate is used in commercial cold packs.
How much energy is required to raise the temperature of 45.0 g of water from 10.5°C to 148.0°C?
Work the problem in 3 steps: 1) Change temperature of 45 g water from 10.5 C to 100 C: q = m c DT = 45 X 4.184 X 89.5 = 16851 J = 16.85 kJ 2) Heat to vaporize 45 g water: 45 g / 18 g/mol = 2.5 mol X 40.7 kJ/mol = 101.75 kJ 3) Heat to raise temperature of steam: q = m c DT = 45 X 2.02 X 48.0 C = 4.36 kJ Add them together to get 123 kJ (3 sig figs because your mass and several other numbers have 3 sig figs).