Answer:
1. 2.510kJ
2. Q = 1.5 kJ
Explanation:
Hello there!
In this case, according to the given information for this calorimetry problem, we can proceed as follows:
1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:
Then, we perform the conversion as follows:
2. Here, we use the general heat equation:
And we plug in the given mass, specific heat and initial and final temperature to obtain:
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<h3>
Answer:</h3>
0.89 J/g°C
<h3>
Explanation:</h3>
Concept tested: Quantity of heat
We are given;
- Mass of the aluminium sample is 120 g
- Quantity of heat absorbed by aluminium sample is 9612 g
- Change in temperature, ΔT = 115°C - 25°C
= 90°C
We are required to calculate the specific heat capacity;
- We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature.
That is;
Q = m × c × ΔT
- Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium.
Specific heat capacity, c = Q ÷ mΔT
= 9612 J ÷ (120 g × 90°C)
= 0.89 J/g°C
Therefore, the specific heat capacity of Aluminium is 0.89 J/g°C
Bronze alloy and porcelain dentures
Answer:
The water phase with the smallest temperature increase when adding 10 kcal of heat is solid ice.
Explanation:
The rest of the statements are incorrect. The density of ice is lower than the density of water. The heat capacity of solid ice is greater almost twice the heat capacity of the liquid water. The heat capacity of vapors is less than heat capacity of liquid.
1) H
2) He
3) Li
4) Be
5) B
6) C
7) N
8) O
9) F
10) Ne
11) Na
12) Mg
13) Al
14) Si
15) P
16) S
17) Cl
18) Ar
19) K
