a liquid with a specific heat of 1900J/Kg° C has 4750 joules of heat energy is added to it. if the temperature increases from 20° to 30°, what is the mass of the liquid
E=mcθ where E is the energy added,m is the mass,c is the specfic heat capacity and θ is the change in temprature. making m subject u get E/cθ=m θ=(30-20)=10 plugging the values we get : [tex] \frac{4750}{1900*10} [/tex] solving it we get the answer that is 0.25kg or 250 grams
