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:
here given is a weight
then force becomes mg
that is F=Mg
=4*9.8
then by using the formula
F=Ma
a=F/M
=4*9.8/9.8
=4
Explanation:
Vocabulary should be, I think:
I. Hypothesis
II. Evidence, data
III. Experiment
What is your question exactly?
Answer:
Specific heat at constant pressure is = 1.005 kJ/kg.K
Specific heat at constant volume is = 0.718 kJ/kg.K
Explanation:
given data
temperature T1 = 50°C
temperature T2 = 80°C
solution
we know energy require to heat the air is express as
for constant pressure and volume
Q = m × c × ΔT ........................1
here m is mass of the gas and c is specific heat of the gas and Δ
T is change in temperature of the gas
here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.
and here at constant pressure Specific heat is greater than the specific heat at constant volume,
so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is
Specific heat at constant pressure is = 1.005 kJ/kg.K
and
Specific heat at constant volume is = 0.718 kJ/kg.K
