How much energy is required to turn 20 gram cube of ice at a temperature of -10 degrees celsius into liquid water at a temperature of 10 degrees celsius

Given that:

Mass (m) = 20 g = 0.02 Kg,

temperature (T₁) = -10°C = -10+273 = 263 K

temperature (T₂) = 10°C = 10+273 = 283 K

Specific heat of water (Cp) = 4.187 KJ/Kg k

We know that Heat transfer (Q) = m. Cp.( T₂ - T₁)

                                                    = 0.02 × 4.187 × (283-263)

                                                 Q = 1.67 KJ

Heat transferred is 1.67 KJ

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

0.19 m/s and 0.45 m/s