M = mass of aluminium = 1.11 kg
= specific heat of aluminium = 900
= initial temperature of aluminium = 78.3 c
m = mass of water = 0.210 kg
= specific heat of water = 4186
= initial temperature of water = 15 c
T = final equilibrium temperature = ?
using conservation of heat
Heat lost by aluminium = heat gained by water
M (
- T) = m
(T -
)
(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)
T = 48.7 c
Answer:
0.0034 sec
Explanation:
L = initial length
T = initial time period = 2.51 s
Time period is given as
L = 1.56392 m
L' = new length
ΔT = Rise in temperature = 142 °C
α = coefficient of linear expansion = 19 x 10⁻⁶ °C
New length due to rise of temperature is given as
L' = L + LαΔT
L' = 1.56392 + (1.56392) (19 x 10⁻⁶) (142)
L' = 1.56814 m
T' = New time period
New time period is given as
T' = 2.5134 sec
Change in time period is given as
ΔT = T' - T
ΔT = 2.5134 - 2.51
ΔT = 0.0034 sec