A silver cube with an edge length of 239 cm and a gold cube...

A silver cube with an edge length of 2.39 cm and a gold cube with an edge length of 2.71 cm are both heated to 82.5 ∘C and placed in 108.0 mL of water at 20.3 ∘C . What is the final temperature of the water when thermal equilibrium is reached Substance Specific heat (J/g·°C) Density (g/cm3) gold 0.1256 19.3 silver 0.2386 10.5 water 4.184 1.00

Transcript

0:00 Hello.

00:01 So in this question we are given that is the edge length of the silver cube.

00:10 We can say that is for ag.

00:12 It is 2 .39 centimeter.

00:16 And edge length for the gold, it is 2 .71 centimeter.

00:23 And the temperature of water, it is 20 .3 degrees celsius.

00:32 So, we need to find out the final temperature of water, then thermal equilibrium is raised at equilibrium.

00:57 So now it is obvious that the heat lost by the gold plus heat lost by the silver, it must be equal to the heat gained by the water.

01:07 Also, we can say that, that is heat lost by the gold.

01:13 Heat, heat lost by heat lost or gained is equal to q, we can say q is equal to m c delta t, where m is the mass c is the specific heat and delta t is a change in the temperature.

01:33 So now let us first calculate the mass of the substances by using the formula, that is density is equal to mass upon.

01:45 Volume and we can say the mass is equal to density into volume so now let us first find out the mass of the silver mass of silver is equal to density into volume and that will be the density of the mass it is we have given all the information in the question and the density here for the silver that is 10 .5 multiply by 2 .39 is the volume so now, see we are talking about the volume here, no? so we have the sides here, edge length.

02:27 That means the volume will be the a cube.

02:30 So therefore here, this will be the cube.

02:33 And on solving this, we will have the mass of silver as 143.

02:46 143 .3 .3 .3 .3 .3 .3 .m.

02:48 325 gram.

02:50 Similarly the mass of gold that will be again the same formula edge length it is here 2 .71 cube multiply by 19 .3 and that will be is equal to the mass of the gold will be 384 .07 gram and the mass of water is equal to volume here volume it is given that is one zero eight aml multiply by the density of water it is one and therefore mass of water it is 108 gram now we will calculate the heat lost by silver and gold and also heat gate by the water now heat lost by the silver heat lost by silver so we have the formula mc delta t.

03:59 So mass here we have calculated that is 143 .3 .325.

04:08 Multiply the specific heat, it is 0 .2386...

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