50 ml of milk at 5°c is added to 200 ml of hot chocolate at 95°c. what do you think the approximate new temperature is?

76.3°C

Considering the fact that;

Specific heat capacity of the hot chocolate = 3.751 kJ/kg C°

Specific heat capacity of milk is 3.93 kJ/(kg.C°)

But;

Heat gained by milk = Heat lost by hot chocolate

If the final temperature is x , then

Heat gained by milk

= 0.05 × 3.93 × (x-5)

= 0.1965x - 0.9825

Heat lost by hot chocolate;

= 0.2 × 3.751 × (95-x)

= 71.269 - 0.7502x

Therefore;

0.1965x - 0.9825 = 71.269 - 0.7502x

0.1965 x + 0.7502x = 71.269 + 0.9825

                  0.9467 x  = 72.2515

                               x = 76.319 °C

Approximately, the new temperature is 76.3°C

your answer would be fusion

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!