Question

2. (15 pts) A 83.5 g sample of a metal alloy is heated to 88.1°C and it is then placed in a coffee-cup calorimeter containing 30.0 g water at 15.0°C. The final temperature of the metal + water is 25.3 °C. Calculate the specific heat of metal alloy, in J/(g°C), assuming no heat escapes to the surroundings or is transferred to the calorimeter. The specific heat of water is 4.184 J/(g°C).

Answer 1

In generally gained = alost we know that heat energy (2) = m.CAT Hot compound will loss the heat Heat lost = Q = m. c; (T-T) To is not compound Initial temperature. cool compound (low Temperature) will gain the heat. Heat gain = a = m2 C₂ (T-T 3 is cool compound Initial temperature. 1 m2 C₂ (T-72) = m c, CT, -T)

Water Initial temperature = 150 alloy Initial temperature = 88.18 :. Alloy will lost the heat Alloy → Boost = Q = mia (T-D) ori = 83-59, 6,=? T= 88.4°C T= Final temperature = Alloy & water = 25.3°C. water will gain the temperature so water again = Q = m Q (T-T2) where given m = 3og, a = specific heat = 4:184 /0 substitute all the values T = 25.3% To= 15% in Qlost = again

mag (T-T2) = mc, (7,-9) sog. (4.181 ° -15%) = 83.59.6 (88.1–25.3) (10.3°C) = 83.59.c, (628°c) 12.92856 J = 52.473.8 gioc. → C, 1297.856 J 5243-8 g.oc G= 0.246 J/g. Oc .: Specific heat of metal alloy = 0.246 3/g.cc