Question

how to do part a, trial 1

Data Analysis Part A. 1. Calculate the specific heat for your metal for each trial. Remember that the heat lost by the metal is equal to the heat gained by the water in the calorimeter and by the calorimeter itself. This can be expressed as - metal (water + Jealorimeter). The specific heats are positive numbers.

Answer 1

Dear Student

- - Part A. (Trial 1) Given: Mass of calorimeter = 40.8713 mass of calorimeter + water - 104.6239 Mass of iron = 64.2599 Initial Temperature of Iron = 100.0°C Initial temperature of water = 24.2°C Final temperature = 31.4°C Calorimeter constant - 22.1 J/g °C To find: Specific heat capacity of Iron Solution: Initial temperature of calorimeter = Initial temperature of water 524.2°C Mass of water = (104.623 -40,871) = 63.7529 We know that I B = m CAT where @= heat gained or lost by substance 1 m = mass of substance (= specific heat capacity of substance AT = Change in temperature Cmater = 4.1 Now, - metal - (Qmater + 0 calorimeter) $ - Mmetal Cmetal 4T metal = Muarter Cuentory A Twater + Mecalori meter 1 (calorimeter 3 - 64.259 Cmetal (31,4 – 100.0) = 63.7523 x 4.186x(31.4 - 24.2) + 40.871 x 22.1 * (31.4-24,2) ater smetal metal 3) • 4408.167 Cmetal = 1921.434 + 6503,393 = 8424.827 4408.167 Ciron = 1.911 5/6°C -metal = 04 LT.827

Theoretical Hence there Ciron = 0.461 5/9°c is error in the observation data.
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