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A 25.0g tin sample was placed in boiling water at 99.5 C until it had the same temperature. Afterwards, the tin sample was placed in room temperature water. If the tin sample lost 399.4J of heat and its final temperature was 24.5 C, what is the specific heat of tin?

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Answer : The specific heat of tin is, 0.213 J/g.K

Explanation :

Formula used :


q=m* c* (T_(final)-T_(initial))

where,

q = amount of heat lost = -399.4 J

c = specific heat capacity of tin = ?

m = mass of tin = 25.0 g


T_(final) = final temperature =
24.5^oC=273+24.5=297.5K


T_(initial) = initial temperature =
99.5^oC=273+99.5=372.5K

Now put all the given values in the above formula, we get:


-399.4J=25.0g* c* (297.5-372.5)K


c=0.213J/g.K

Therefore, the specific heat of tin is, 0.213 J/g.K

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