Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
You can put a known amount sodium into some sort of time release mechanism such as a pill made from soluble material. Then you can place the sodium into a calorimeter with a known mass of water and record the temperature change the water undergoes during the reaction. Then you can use the equation q(water)=m(water)c(water)ΔT to find the amount of heat absorbed by the water. since the amount of heat absorbed by the water is the amount of heat released from the sodium, q(sodium)=-q(water). Than you can use the equation q(sodium)=m(sodium)c(sodium)ΔT and solve for c(sodium)
I hope this helps and feel free to ask about anything that was unclear in the comments.
<span>1. Fill a beaker or graduated cylinder with enough water to completely immerse the sphere in. 2. Record the baseline initial measurement. 3. Drop the sphere in. 4 <span>Record final measurement.</span></span>
Answer:
<h2>0.06 % </h2>
Explanation:
The percentage error of a certain measurement can be found by using the formula
From the question
error = 500 - 499.7 = 0.3
actual volume = 500 mL
We have
We have the final answer as
<h3>0.06 % </h3>
Hope this helps you
Answer:
B. Thicker layer of blubber
Explanation:
For the leopard seals to survives this harsh environment, it must have a thicker layer of blubber
Blubber is a thick layer of fat. It is called the adipose tissues and found in most marine organisms.
- This layer helps in insulating the body against heat loss.
- By so doing, the animal is able to conserve internal heat.
- They have low thermal conductivity and do not easily lose heat or gain heat as such.
