10. How much energy must be added to 2800 g of gold at its melting point...
10.
How much energy must be added to 2800 g of gold at its melting
point of 1063 deg. C in order to change it from a solid to a liquid
at 1121.0 deg. C?
Answer: Last
Answer: 1.7676E+05 J
Not yet correct, tries 3/20
I NEED THE UNITS TO BE CORRECT PLEASE. THEY'RE BEING USED ON CAPA. ALSO SHOW ALL WORK. THANKS.
Thermal Properties of Some Materials
| Material | alpha | beta | csolid | Tmelt | Lfusion | cliquid | Tboil | Lvaporize | cgas | k |
| K-1 | K-1 | J kg-1 K-1 | oC | J kg-1 | J kg-1 K-1 | oC | J kg-1 | J kg-1 K-1 | J s-1 m-1 K-1 | |
| Air | -- | -- | -- | -- | -- | -- | -- | -- | -- | 0.0256 |
| Aluminum | 26 x 10-6 | 75 x 10-6 | 900 | 660 | 380 x 103 | 900 | 2450 | 11400 x 103 | 900 | 240 |
| Ammonia | -- | -- | 4700 | -77.8 | 3320 x 103 | 4700 | -33.4 | 1371 x 103 | 2060 | -- |
| Brass | 19 x 10-6 C-1 | 56 x 10-6 C-1 | -- | -- | -- | -- | -- | -- | -- | 110 |
| Concrete | -- | -- | -- | -- | -- | -- | -- | -- | -- | 1.1 |
| Copper | 17 x 10-6 C-1 | 51 x 10-6 C-1 | 378 | 1083 | 134 x 103 | 378 | 2565 | 5069 x 103 | 378 | 390 |
| Ethyl Alcohol | -- | 1100 x 10-6 C-1 | 2450 | -114.4 | 1080 x 103 | 2450 | 78.3 | 855 x 103 | 1680 | -- |
| Gasoline | 0 x 10-6 C-1 | 950 x 10-6 C-1 | -- | -- | -- | -- | -- | -- | -- | -- |
| Glass | 9 x 10-6 C-1 | 27 x 10-6 C-1 | -- | -- | -- | -- | -- | -- | -- | 0.8 |
| Ethelyne Glycol | 0 x 10-6 C-1 | 570 x 10-6 C-1 | -- | -- | -- | -- | -- | -- | -- | -- |
| Gold | 14 x 10-6 C-1 | 42 x 10-6 C-1 | 129 | 1063 | 64.5 x 103 | 129 | 2660 | 1578 x 103 | 129 | -- |
| Goose Down | -- | -- | -- | -- | -- | -- | -- | -- | -- | 0.025 |
| Hydrogen | -- | -- | 967 | -259.3 | 58.6 x 103 | 967 | -252.9 | 452 x 103 | 1432 | -- |
| Invar | 0.9 x 10-6 C-1 | 2.7 x 10-6 C-1 | -- | -- | -- | -- | -- | -- | -- | -- |
| Iron | 12 x 10-6 C-1 | 35 x 10-6 C-1 | -- | -- | -- | -- | -- | -- | -- | 79 |
| Kerosene | 0 x 10-6 C-1 | 990 x 10-6 C-1 | -- | -- | -- | -- | -- | -- | -- | -- |
| Lead | 29 x 10-6 C-1 | 87 x 10-6 C-1 | 128 | 327 | 24.5 x 103 | 128 | 1750 | 871 x 103 | 128 | 35 |
| Nitrogen | -- | -- | 2042 | -210 | 25.5 x 103 | 2042 | -195.8 | 201 x 103 | 1040 | -- |
| Oxygen | -- | -- | 1669 | -218.8 | 13.8 x 103 | 1669 | -183 | 213 x 103 | 919 | -- |
| Pyrex | 3 x 10-6 C-1 | 9 x 10-6 C-1 | -- | -- | -- | -- | -- | -- | -- | -- |
| Silver | 18 x 10-6 C-1 | 18 x 10-6 C-1 | -- | -- | -- | -- | -- | -- | -- | 420 |
| Steel | 12 x 10-6 C-1 | 35 x 10-6 C-1 | 449 | 1538 | -- | -- | 2862 | -- | -- | 79 |
| Styrofoam | -- | -- | -- | -- | -- | -- | -- | -- | -- | 0.01 |
| Water | -- | 214 x 10-6 C-1 | 211 | 0 | 334 x 103 | 4186 | 100 | 2256 x 103 | 2080 | 0.6 |
| Wood | -- | -- | -- | -- | -- | -- | -- | -- | -- | 0.15 |
| Wool | -- | -- | -- | -- | -- | -- | -- | -- | -- | 0.04 |
Homework Answers
Mass of gold M=2800g
Initial temperature T1=1063 °C
finial temperature T2=1121°C
Specific heat s=129
This concept belongs thermal properties of meter
So heat energy required for specific heat
Q=MS(t2-t1)
=2800×129(1121-1063)
=20949600J
=209.5×10^5 J
Therefore the heat energy required =209.5×10^5 J
Add Answer to:
10.
How much energy must be added to 2800 g of gold at its melting
point...
10. [2pt] How much energy must be added to 2800 g of gold at its melting...
10. [2pt] How much energy must be added to 2800 g of gold at its melting point of 1063 deg. C in order to change it from a solid to a liquid at 1121.0 deg. C? Answer: Last Answer: 4.64E+05 J Not yet correct, tries 1/20
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