Question

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

Answer 1

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.

Answer 2

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.