A 50 g ice cube floats in 195 g of water in a 100 g copper cup; all are at a temperature of 0°C. A piece of lead at 96°C is dropped into the cup, and the final equilibrium temperature is 12°C. What is the mass of the lead?

The mass of the lead will be "1.127 kg".

Explanation:

The given values are:

(Ice) m₁ = 50 g i.e.,

0.050 kg

(Water) m₂ = 195 g i.e.,

0.190 kg

(Copper cup) m₃ = 100 g i.e.,

0.100 kg

m₁, m₂ and m₃ at temperature,

t₁ = 0°C

Temperature of lead,

t₂ = 96°C

Temperature of Final equilibrium,

t₃ = 12°C

Let m₄ be the mass of the lead.

On applying formula, we get

⇒  

On putting the estimated values, we get

⇒  

⇒  

⇒  

⇒  

⇒  

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