Answer:
a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
b) T = 295.37 K
Explanation:
Given;
Initial temperature of tea T1 = 31 C
Initial temperature of ice T2 = 0 C
Mass of tea m1 = 0.89 kg
Mass of ice m2 = 0.075kg
The heat capacity of both water and tea c = 4186 J/(kg⋅K)
the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg
And T = the final temperature of the mixture
Heat loss by tea = heat gained by ice
m1c∆T1 = m2c∆T2 + m2Lf
m1c(T1-T) = m2c(T-T2) + m2Lf
m1cT1 - m1cT = m2cT - m2cT2 + m2Lf
m1cT + m2cT = m1cT1 + m2cT2 - m2Lf
T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
Substituting the values;
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)
T = 22.37 °C
T = 273 + 22.37 K
T = 295.37 K
Answer:
0.002925 m
Explanation:
Lt = LO(1 +α Δt ) here Lt is total length Lo is original length α is coefficient of linear expansion and Δt is change in temperature
<h2>for aluminium</h2>
α=25×10^-6
Lt = 5(1+25×10^-6×(70-20))
Lt = 5 (1+25×10^-6×50)
Lt = 5 ( 1+0.00125)
Lt = 5×1.00125
Lt =5.00625 m
<h2>for nickel </h2>
α=13.3×10^-6
Lt =5(1+13.3×10^-6×50)
Lt = 5(1+0.000665)
Lt =5.003325 m
hence difference in length =5.00625-5.003325
= 0.002925 m
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