: Activity: A liquid of unknown specific heat at a temperature of 20°C wa: 80°C in a well-insulated container. The final temperature was measure combined mass of the two liquids was measured to be 240g. In a se both liquids at the same initial temperatures, 20 g less of the liquid of was poured into the same amount of water as before. This time the ec was found to be 52°C. Determine the specific heat of the liquid. The s 4187 J/Kg°C or 1 kcal/kg°C.

Specific heat capacity of unknown liquid is given as,

°

Explanation:

Let the mass of unknown liquid is "m"

Now when unknown liquid is mixed with water then heat given by water is equal to the heat absorbed by unknown liquid

So we have

Msliq ( 50 - 20 ) = (240 - m) (4.2)(80-50)

Now again we did the same experiment but this time the mass of unknown liquid is "m - 20"

so we have

(m-20)Sliq(52-20)=(220-m)(4.2)(80-52)

now from above two equations  

m(30)/(m-20)32 = (240-m)30/(220-m)/28

Now we have

30m * 28 (220 - m) = 32 ( m- 20) * 30 ( 240 - m)

so we have

m = 84.2g

Now we have

Sliq = 240-84.2/84.2 (4.2)

Sliq = 7.77J/g°C

i believe the answer is electric potential.

explanation:

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y