Question

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

Answer 1

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.

Answer 2

Answer: The temperature increase will be 31.70°C.

Step-by-step 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* 4.186J/g^oC* \Delta T\\\\\Delta T=31.70^oC`

Hence, the temperature increase will be 31.70°C.