Answer:
Final temperature = T₂ = 155.43 °C
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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
Given data:
Mass of coin = 4.50 g
Heat absorbed = 54 cal
Initial temperature = 25 °C
Specific heat of copper = 0.092 cal/g °C
Final temperature = ?
Solution:
Q = m.c. ΔT
ΔT = T₂ -T₁
Q = m.c. T₂ -T₁
54 cal = 4.50 g × 0.092 cal/g °C × T₂ -25 °C
54 cal = 0.414 cal/ °C × T₂ -25 °C
54 cal /0.414 cal/ °C = T₂ -25 °C
130.43 °C = T₂ -25 °C
130.43 °C + 25 °C = T₂
155.43 °C = T₂
Answer:
Q = 1461.6 J
Explanation:
Given data:
Mass of ice = 36 g
Initial temperature = -20°C
Final temperature = 0°C
Amount of heat absorbed = ?
Solution:
specific heat capacity of ice is 2.03 j/g.°C
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 = T2 - T1
ΔT = 0°C - (-20°C)
ΔT = 20°C
Q = 36 g ×2.03 j/g.°C×20°C
Q = 1461.6 J
8.03 solutions report is described below.
Explanation:
8.03 Solutions Lab Report
In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.
Pre-lab Questions:
In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.
Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).
Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.
Answer:
2-4 mm height of capillary tube.
Explanation:
Sample should be around 2-4 mm in height.
It should be packed well so that it does not have air packets that caues the lowering of melting point.
If you take greater amount, then there will be needed more heat, resulting a wide range of melting point.
