General Chemistry I CHEM-1030 Laboratory Experiment No. 10 Specific Heat Prestud
ID: 506111 • Letter: G
Question
General Chemistry I CHEM-1030 Laboratory Experiment No. 10 Specific Heat Prestudy Calculate the heat needed to raise the temperature of 2789.8 g of water from 23.8 eC to 89.8 oC. (2 points) What is the specific heat of a metal if 65.289 g of the metal requires 939.0 joules for a 19.0 eC (2 points) temperature change? A 63.201 g sample of an unknown metal at 99.9 °C is mixed with 182.43 g of water at 21.6 oC. The resulting mixture reaches a temperature of 24.9 aC. Calculate the specific heat of the metal by solving Equation 8 for smetal. Equation 8, which is derived from Equations 5, 6 and 7, assumes the heat gained by the container (calorimeter) is negligible in comparison to the heat gained by the points) Water. (Answer question #4 on the reverse side of this namn)Explanation / Answer
1.
Heat required (q) to raise the temperature from 23.8 °C to 89.8 °C (boiling).
Q = (mass) (specific heat) x (temp. Difference)
= m x C x T
Mass = 2789.8 g
Specific heat of water = 4.184 J / g°C
T = 89.8 °C – 23.8 °C = 66 °C.
So,
Q = (2789.8 g) x (4.184 J / g°C) x (66 °C)
= 770386 J
= 770.39 kJ
2.
Heat required (Q) to raise the temperature from Ti to Tf.
Q = (mass) (specific heat) x (Tf - Ti)
= m x C x T
Here,
Heat, Q = 939 J
mass, m = 65.289 g
Change in temperature, T = Tf - Ti = 19 oC
Specific heat of metal, C = ?
So,
939 J = (65.289 g) x (C) x (19 °C)
C = (939 J) / [(65.289 g) x (19 °C)]
= 0.757 J/g °C
3.
Heat required (Q) to raise the temperature from Ti to Tf.
Q = (mass) (specific heat) x (Tf - Ti)
= m x C x T
specific heat metal (Cm) = ?
Tm = 99.9 oC – 24.9 oC = 75.0 oC
Mm = 63.201 g
specific heat (CH2O) = 4.184 J / g°C
TH2O = 24.9 °C – 21.6 °C = 3.3 °C
MH2O = 182.43 g
Now,
Heat released = Heat absorbed
specific heat (Cm) x Tm x Mm = specific heat (CH2O) x TH2O x MH2O
specific heat (Cm) x (75.0 oC) x (63.201 g) = (4.184 J/g°C) x (3.3 °C) x (182.43 g)
specific heat (Cm) = [(4.184 J/g°C) x (3.3 °C) x (182.43 g)] / [(75.0 oC) x (63.201 g)]
specific heat (Cm) = 0.531 J/g°C
Image 1, all parts are done. Image 2 is incomplete question or no data given.
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