When 0.187 g of benzene. C_6H_6, is burned in a bomb calorimeter, the surroundin

Question

When 0.187 g of benzene. C_6H_6, is burned in a bomb calorimeter, the surrounding water bath rises in temperature by 7.48 degree C. Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate combustion energies (Delta E) for benzene Instant cold packs used to treat athletic injuries contain solid NH_4NO_3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, lowering the temperature because of the exothermic reaction What is the final temperature m a squeezed cold pack that contains 50 0 p. of NH_4NO_3 dissolved in 125 ml of water? Assume a specific heat of 4.18 J/(g'C) for the solution, an initial temperature of 25.0'C, and no heat transfer between the cold pack and the environment When a 25 0 g piece of metal absorbs 50 0 J of heat, its temperature increases by 15 5'C What made from Silver

Explanation / Answer

1. Heat transferred to water= Mass of water* specific heat of water* temperature rise (since heat capacity of   calorimeter can be neglected)

= 250*4.18*7.48=7817 joules

This heat is released due to combustion of 0.187 gm of Benzene

Hence combustion energy= 7817/0.187 j/gm=41802.14 j/g= 41.802 Kj/g

Moles of Benzene = mass/Molecular weight= 0.187/78=0.002397

Combustion energy= 7817/0.002397=3260567 j/mol=3260.567 Kj/mol

2.

Mass of the solution =mass of NH4NO3+ mass of water= 50+125*1 =175

(water density is assumed as 1 g/cc)

Specific heat= 4.18 j/g.deg.c

Heat lost= mass*specific heat* temperature difference

25.7*1000= 175*4.18*(25-T) T is the final temperature

T= -10.133 deg.c

3.

Heat abosrbed=mass* specific heat* temperature difference

50= 25*15.5*Cp ( Cp =specific heat of metal)

Cp =0.129 j/g.deg.c

The metal is gold

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