Need help on these two questions uranium metal specific heal g c at 200.0 \"C is

Question

Need help on these two questions

uranium metal specific heal g c at 200.0 "C is dropped into 1.00 L of deuterium oxide 4.21119 i) at 25.5 of "heavy water specific heat ture of the uranium and the deuterium oxide The final temperat mixture is 2B.5 C. Given the densities of uranium (19.05 the side of the uranium cube? To solve for this, you will have to ne the mass of uranium dropped the heat lost by the urankum the heat gained by the water using the equation we discussed in b. Use the density equation to calculate the volume e Determine the length by using the equation that relates volume and length or a cube 4. 175.0 g pure Ho was placed n a constant pressure 0 C. 080 g pure Hiso, lalso 10.0 was added, stirred and the temperature nose to 198 What mass increased in temperature? Icl What was the chemical (d What do we sssume about the specific heat? Calculate the energy change for the reaction

Explanation / Answer

Let m =mass of uranium metal

Heat lost by uranium metal =heat gained by heavy water ( under adiabatic conditions)

Heat gained or lost= mass* specific heat* temperature difference

m*0.117*(200-28.5)= 1L*1.1*1000g/L * 4.211* (28.5-25.5)

m= 164.5 gm

Density of uranium =19.05 g/cc

Volumeof uranium = 164.5/19.05=8.63cc

Volume of cube= l3= 8.63 ( l is length of cube), l =2.05cm

b)

9.8 gms of H2SO4 increased the temperature

Delt= 19.8-10= 9.8 deg.c

The reaction is H2SO4+H2O....> HSO4- + H3O+

Once mixed (sulfuric acid is 9.8 gms compared to 175 gm of water, hence specific heat of mixture can be assumed same as that of water. ( 4.18 j/g.deg.c)

Energy change= Total mass ( 175+9.8)* 4.18 j/g.deg.c (19.8-10)= 7570.147 joules

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