Please help need answer asp
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Answer:
e% = 3.4%
Explanation:
This is a calorimetry problem where the heat released equals the heat absorbed
m (T₀ - T_f) = M c_{e2} (T₁ - T_f)
Index 1 refers to water and index 2 to metal, in this case it asks for the specific heat of the metal (c_{e2})
c_{e2} = m / M c_{e1} (T_f -T₀) / (T₁ - T_f)
Let's calculate
c_{e} = 60/100 4.19 (24-20) / (100-24)
c_{e2} = 0.1323 j / gC
This metal is possibly lead, which is its specific heat is 0.128 J / gC
The percentage error is
e% = (c_{e2} - 0.128) /0.128 100
e% = 3.4%