The specific heat capacity of the metal given the data from the question is 0.66 J/gºC
<h3>Data obtained from the question</h3>
- Mass of metal (M) = 76 g
- Temperature of metal (T) = 96 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 24.5 °C
- Equilibrium temperature (Tₑ) = 31 °C
- Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
- Specific heat capacity of metal (C) =?
<h3>How to determine the specific heat capacity of the metal</h3>
The specific heat capacity of the sample of the metal can be obtained as follow:
Heat loss = Heat gain
MC(M –Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
76 × C × (96 – 31) = 120 × 4.184 × (31 – 24.5)
C × 4940 = 3263.52
Divide both side by 4940
C = 3263.52 / 4940
C = 0.66 J/gºC
Learn more about heat transfer:
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Answer:
Mass of 205 mL of the liquid is 164 g
Explanation:
We know, 
Here density of liquid is 0.798 g/mL and volume of liquid is 205 mL
So, mass of liquid = 
= 
= 164 g
hence mass of 205 mL of the liquid is 164 g.
Answer:
The energies of combustion (per gram) for hydrogen and methane are as follows: Methane = 82.5 kJ/g; Hydrogen = 162 kJ/g
<em>Note: The question is incomplete. The complete question is given below:</em>
To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/℃. When a 1.00-g sample of methane gas burned with
<em>excess oxygen in the calorimeter, the temperature increased by 7.3℃. When a 1.00 g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3°C. Compare the energies of combustion (per gram) for hydrogen and methane.</em>
Explanation:
From the equation of the first law of thermodynamics, ΔU = Q + W
Since there is no expansion work in the bomb calorimeter, ΔU = Q
But Q = CΔT
where C is heat capacity of the bomb calorimeter = 11.3
kJ/ºC; ΔT = temperature change
For combustion of methane gas:
Q per gram = (
11.3
kJ/ºC * 7.3°C)/1.0g
Q = 83 kJ/g
For combustion of hydrogen gas:
Q per gram = (
11.3
kJ/ºC * 14.3°C)/1.0g
Q = 162 kJ/g
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