Question

How do I find the specific heat (experimental) and percent Error?

The calorimeter is a styrofoam cup that weighs 19.9 g

	Aluminum	Copper	Iron
Mass of Sample (g)	18.4	57.7	48.7
Mass of Calorimeter and cold water (g)	268.1 g	267.3 g	267.1 g
Mass of Cold Water (g)	248.2 g	247.4 g	247.2 g
Initial Temperature of Cold Water	16.6 °C	18 °C	18 °C
Temperature of boiling water	100 °C	100 °C	100 °C
Equillibrium Temperature	17 °C	19.5 °C	21 °C
Specific Heat (experimental)	?	?	?
Specific Heat (standard)	0.215 cal/(g°C)	0.0924 cal/(g°C)	0.107 cal/(g°C)
Percent Error (%)	?	?	?

Answer 1

0 For Aluminium Mass of the sample (ms) 18.40 Mass of the calorimeter and cold water 268. g. Mars a cold water (mw) = 248.2g Intial temperature of cold water &- 16.6°C Temperature of boiling water -200°c Equilibrium temperature -17°c We know, heat energi og insat where im' is the mass 's' is specific heat and 'At is the change of temperature Now, I metal - Iwater mg xls) 2 ht im HE - mw x(s) water + (at) water – o How specific heat of water swater = 4.284 Ugoci And initially the metal heated up by immersion in boiling water so we can take the initial temperature of metal is tooe So, from o (m) metal (3) me lak * (at) metal water water het water (18.4g) x (S) meta X (100-17) (= (248.2g) = (4.28 4 j goc *(17-16-6)* | (s) metal = 0.272 yg loci (experimental) Standard value of specific heat for Aluminum given by 0.215 cal/ge. Expermentaly Smetal = 0: 272 yg local .. 0.272 4184 =0.065 cal/g °C

(standard value Experimental percent Error = 0.215 0.065 noy. Percent en 0.215 = 69.764) value) : ② For Copper Standard value] Given * 100 Mass of copper Y. = 57.7g Mass of water = 247.49 al Temperature of cold water = 18°C Initial temperature of copper - 100°c Equilibrium temperature = 19.5°c specific heat of water - 1 cal/gie I copper - I water (m) copper & (s) copper & (A) opper: (m.sat) water (57. g* (S) copper* (100-19.5) - (247.44) Xcal /ae) x (19.5-18) Scopper = 0.0799 cal/zel Standard value (S)e opper - 0.0924 call que : Percent. Error 0.0924 -0.0799 x 100%. 0.0924 = 113.53 %) ③ For Iron Given mass of Iron = 48.7g Mass of water = 247.2g Initial temperature of cold water 18°c Equillibrinm temperature = 21°C & Iron = 9 water (m) Iron & Saron Xat) = (m.s.at) water (48.7g) x Stronx (100-21) = (247.2g) x 2 cal/gie x (21-15 Stron = 0.193 cal/gel Standard value of specific heat of Iron = 0.10 7 calle percent error = 0.19.3-0.107 x100%! 0.107 = 180.3%)