l0 at 8 AM CDT is allowed, with a one time 10% penalty to submitted score. Use t

Question

l0 at 8 AM CDT is allowed, with a one time 10% penalty to submitted score. Use the References to access important values If needed for this question In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal. She heats 18.2 grams of manesium to 99.22°C and then drops it into a cup containing 76.0 grams of water at 21.60 C. She measures the final temperature to be 25.75°c Assuming that all of the heat is transferred to the water, she calculates the specific heat of magnesium to be J/g C. Submit Answer offee Cup Calorimetry Heat of Solution :This is group attempt 1 of 5 Autosaved at 502 PM MacBook Air 0

Explanation / Answer

Heat absorbed by water = H1 = mass of water * specfic heat of water * (Final temperature - Initial temperature) = 76 g * 4.18 J/ g C * ( 25.75-21.60 ) C = 1318.372 J

let us assume that specfic heat of magnesium is s J/ g C.

Heat released by magnesium = - H2 = - mass ofmagnesium* specfic heat of magnesium * (Final temperature - Initial temperature) = - 18.2 g *s J/ g C * ( 25.75-99.22 ) C = 1336.79* s J

At thermal equilibrium:

H1 = H2

1318.372 = 1336.79* s

So, s = (1318.372/ 1336.79) = 0.986

Answer: specfic heat of magnesium is 0.986 J/ g C.

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