calorimetry?

Question

in a experiment 200kg of aluminium (c=880 J/kgC0 at 100C is dropped into 256kg of kerosene at 15C. the mixture reaches a temperature of 50C. what is the specific heat of kerosene? show working out

Answer 1

you write the absolute value of energy lost by Al will be equal to that received by kerosene

Use formula dE = mc (tf -ti)
tf = final temperature , ti initial temperature , specific heat

so for Al dE =200 *880* (100-50) =8.8 10^6 J
fo kerosene dE = 256*c * (50-15)= 256*35*c J

you find c=982J/kg