If a 375 ml sample of water was cooled from 37.5 degrees celsius to 0 degrees celsius , how much heat was lost (in joules)

Explanation:

It is given that volume is 375 mL and is 37.5 degree celsius.

First, we will calculate the mass of sample of water as it is known that density of water is 1 .

Hence,          Density =

                   1 =

                           mass = 375 g

Therefore, formula to calculate heat loss will be as follows.

                         q = mC

where          q = heat lost or absorbed

                    m = mass

                    C = specific heat of water = 4.186 J/g

  = change in temperature

Hence, we will calculate value of heat lost as follows.

                      q = mC

                         = 375 g × 4.186 J/g × ( 0 - 37.5) degree celsius

                          = - 58865.62 joules

Thus, we can conclude that heat lost is -58865.62 joules.

given the reactants, h2so4 and koh, we are asked to determine the product(s) of the reaction. since h2so4 is an acid and koh is a base. this is a neutralization reaction, thus, the products are salt and water. 

so, h+ from h2so4 will react with oh- from koh forming h2o. k+ will also react with so4 2- forming k2so4. therefore, the balanced reaction will be:

h2so4 + 2koh > k2so4 + 2h2o

explanation: