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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

conduction, convection, radiation