1.00kg of ice at -24 degrees Celsius is placed in contact with a 1.00kg block of a metal at 5.00 degrees Celsius. They come to equilibrium at -8.88 degrees Celsius. What is the specific heat of the metal?

2178 J/(kg*C)

Explanation:

Trust me bro

C₂ = 2.22 KJ/kg °C

Explanation:

Since, both objects are in thermal contact. Therefore, the law of conservation of energy tells us that:

where,

m₁ = mass of ice = 1 kg

C₁ = specific heat of ice = 2.04 KJ/kg.°C

ΔT₁ = Change in Temperature of Ice = -8.88°C - (-24°C) = 15.12°C

m₂ = mass of metal block = 1 kg

C₂ = specific heat of metal = ?

ΔT₂ = Change in Temperature of Metal Block = 5°C - (-8.88°C) = 13.88°C

Therefore, using these values in the equation, we get:

C₂ = 2.22 KJ/kg °C

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer