The specific heat of the metal, given the data from the question is 0.60 J/gºC
<h3>Data obtained from the question </h3>
The following data were obtained from the question:
- Mass of metal (M) = 74 g
- Temperature of metal (T) = 94 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 26.5 °C
- Equilibrium temperature (Tₑ) = 32 °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 gold can be obtained as follow:
According to the law of conservation of energy, we have:
Heat loss = Heat gain
MC(T –Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)
74 × C(94 – 32) = 120 × 4.184 (32 – 26.5)
C × 4588 = 2761.44
Divide both side by 4588
C = 2761.44 / 4588
C = 0.60 J/gºC
Thus, the specific heat capacity of the metal is 0.60 J/gºC
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Answer:
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Explanation:
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The frequency of a 4,600 cm wave is 652 Hz
Frequency is the number of waves that pass a fixed point in unit time
Here given data is
Wave = 4,600 cm = 4.6 × 10⁸ nm
We have to calculated frequency = ?
Frequency =v = c/λ
Frequency = 3×10⁸m / 4.6 × 10⁸ nm
Frequency = 652 Hz
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Answer:
3.46
⋅
10
−
19
J
Explanation:
The energy of a photon is proportional to its frequency,
Answer: The enthalpy change during this reaction is 77505.56 J.
Explanation:
Given: 
, 
Mass = 254 g, Specific heat = 
Formula used to calculate the enthalpy change is as follows.
where,
q = enthalpy change
m = mass of substance
C = specific heat capacity
= initial temperature
= final temperature
Substitute the values into above formula as follows.
Thus, we can conclude that the enthalpy change during this reaction is 77505.56 J.
