How much heat is required to heat 1.8 kg of water from 8.4oC to 17.3oC? The heat capacity of water is 4.1813 J/g*oC

Q = 66984.43 J

Explanation:

Given data:

Mass of water = 1.8 Kg (1.8Kg×1000 g/1 Kg =1800 g)

Initial temperature = 8.4°C

Final temperature = 17.3°C

Specific heat capacity of water = 4.1813 J/g.°C

Heat required = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 17.3°C - 8.4°C

ΔT = 8.9°C

Q = 1800 g ×  4.1813 J/g.°C ×8.9°C

Q = 66984.43 J

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

answer: i dont know the answer but this might

explanation:

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