How much would the temperature of 275 g of water increase if 36.5 kj of heat were added?

Chemistry

2 answers:

Andrew [12]3 years ago

8 0

Answer: The temperature increase will be 31.70°C.

Explanation:

To calculate the increase in the temperature of the system, we use the equation:

$q=mc\Delta T$

where,

q = Heat absorbed = 36.5 kJ = 36500J

m = Mass of water = 275 g

c = Specific heat capacity of water = $4.186J/g^oC$

$\Delta T$ = change in temperature = ? °C

Putting values in above equation, we get:

$36500J=275\times 4.186J/g^oC\times \Delta T\\\\\Delta T=31.70^oC$

Hence, the temperature increase will be 31.70°C.

BlackZzzverrR [31]3 years ago

8 0

The solution is found using the formula change in Temperature = heat supplied/ (mass of substance x specific heat). The solution would be change in temperature = 36.5kJ/(275 g x 4.184 J/g) = 31.7 degrees.

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A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new

Artyom0805 [142]

Answer: There are now 2.07 moles of gas in the flask.

Explanation:

$PV=nRT$

P= Pressure of the gas = 697 mmHg = 0.92 atm (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

$V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L$

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

$PV=nRT$

P= Pressure of the gas = 775 mmHg = 1.02 atm (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

$n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles$