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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Violet, indigo, blue, green, yellow, orange, red
Answer:
Final mass = 159.5 g
Final temperature = 10 C
Final density = 1.00 g/ml
Explanation:
<u>Given:</u>
Beaker 1:
Mass of water = 44.3 g
Temperature = 10 C
Beaker 2:
Mass of water = 115.2 g
Temperature = 10 C
Density of water at 10C = 1.00 g/ml
<u>To determine:</u>
The final mass, temperature and density of water
<u>Calculation:</u>
Since there is no change in temperature, the final temperature will be 10 C
Density of a substance is an intensive property i.e. it is independent of the mass. Hence the density of water will remain constant i.e. 1.00 g/ml
34g C * ( 1 mol / 12.0107 ) * ( 1 mol H2 / 1 mol C ) * ( <span>2.01588 g / 1 mol H2 ) = 5.70657164028741 g H2 = 5.7 g H2
Convert grams of C to moles of C using the given amount of grams and the molar mass ( 12.0107 g/mol ).
Gather the mole ratio from the coefficients in the balanced equation and multiply by the ratio.
Convert moles of H2 to grams of H2 </span> using the given amount of grams and the molar mass ( 2.01588 g/mol )<span>.
Revise your answer to have the correct number of significant figures. </span>
Answer: Location 3 is warmer than location 2 and 1
Explanation:
