10. How much energy must be added to 2800 g of gold at its melting point of 1063

Question

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
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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

Explanation / Answer

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

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