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3 years ago

How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat

Chemistry

1 answer:

inna [77]3 years ago

8 0

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

=670*4.1*40.3

=110704.1

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Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

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$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

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$P_1$ = initial pressure of gas = 3 atm

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Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{3atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}$

$P_2=3.918atm$

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