In the laborator a c e cu ca or ter, or constant ressure ca or meter is equent u

Question

In the laborator a c e cu ca or ter, or constant ressure ca or meter is equent used to deter une e spe iche a so d or o me sure e ener o a·lution hase A student heats 62.24 grams of tungsten to 98.89 °C and then drops it into a cup containing 80.79 grams of water at 23.20 °C. She measures the final temperature to be 24.94°C The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.52 JPC Assuming that no heat is lost to the surroundings calculate the specific heat of tungsten. Specific Heat (W) = action. ligC.

Explanation / Answer

1.Tungsten is at higher temperature and looses heat while calorimeter and water gain heat. for adiabatic system

Heat lost by tungsten= heat gained by calorimeter+ heat gained by water

Mass of tungsten*specific heat of tungsten* temperature drop = heat capacity of calorimeter* temperature difference + mass of water* specific heat of water* temperature change

Let CP= specific heat of metal, specific heat of water= 4.184 J/gm.deg.c

62.24*Cp*(98.89-24.94) =1.52*(24.94-23.2)+ 80.79*4.184*(24.94-23.2)

CP = 0.128 j/gm.deg.c

2.

Nickel looses heat and water and calorimeter gains heat

Specific heat of nickel = 0.5 J/gm.deg.c

C= heat capacity of calorimeter

Specific heat of water= 4.184 J/gm.deg.c

Heat lost be nickel= heat gained by water+ heat gained by calorimeter

Mass of nickel* specific heat* temperature difference = mass of water* specific heat of water* temperature difference+ heat capacity of calorimeter* temperature difference

98.24*0.5*(98.74-30.39)= 82.46*4.184*(30.39-21.79)+C*(30.39-21.79)

3357.352= 2967.1+C*8.6

C= 45.37 J/deg.c

3.

Heat released during combustion = heat gained by water+ heat gained by calorimeter

= mass of water* specific heat of water* rise in temperature + heat capacity of calorimeter* temperature rise

=1.198*1000*4.184*(28.12-25.23+768.5*(28.12-25.23)

=16707 joules

This is heat released by 1.369 gm of maleic acid ( C4H4O4)

Moles   = mass/moles = 1.369/116=0.0118

Heat released/mole= 16707/0.0118 j/mole=1415632 J/mole =1415.632 Kj/mole

4. heat added= mass of silver* specific heat of silver*change in temperature

specific heat of silver =0.24 J/gm.deg.c, t= initial temperature

42=11.9*0.24*(37.9-t)

t=23.19 deg.c

5. specific heat of copper = 0.385 J/gm.deg.c, t= final temperature of copper

heat added= heat taken by copper

75.1= 10.6*0.385*(t-20.9), t= 39.3 deg.c

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.