when 20 grams of a salt is dissolved in 95 grams of water, the temperature of the solution decreases from 40oc to 0oc. 880 calories heat is absorbed, the specific heat of the solution is 1.1 Cal/goc.
delta t = T f - T i
= 0 - 40
= -40 degrees Celsius
c = 1.1 Cal/g degrees Celsius
m = 20 grams
heat absorbed (q) = mc(delta)t
q = 20 * 1.1 * -40 degrees Celsius
q = 880 Cal
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