A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at
1 answer:
Answer:
Explanation:
Given
mass of lead piece
mass of water in calorimeter
Initial temperature of water
Initial temperature of lead piece
we know heat capacity of lead and water are and
respectively
Let us take be the final temperature of the system
Conserving energy
heat lost by lead=heat gained by water
You might be interested in
Answer:
v₀ = 280.6 m / s
Explanation:
we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,
We write the mechanical energy when the shock has passed the bodies
Em₀ = K = ½ (m + M) v²
We write the mechanical energy when the spring is in maximum compression
½ (m + M) v² = ½ k x²
Let's calculate the system speed
v = √ [k x² / (m + M)]
v = √[152 ×0.78² / (0.012 +0.109) ]
v = 27.65 m / s
This is the speed of the bullet + Block system
Now let's use the moment to solve the shock
Before the crash
p₀ = m v₀
After the crash
The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved
m v₀ = (m + M) v
v₀ = v (m + M) / m
let's calculate
v₀ = 27.83 (0.012 +0.109) /0.012
v₀ = 280.6 m / s
We can salve the problem by using the formula:
where F is the force applied, k is the spring constant and x is the stretching of the spring.
From the first situation we can calculate the spring constant, which is given by the ratio between the force applied and the stretching of the spring:
By using the value of the spring constant we calculated in the first step, we can calculate the new stretching of the spring when a force of 33 N is applied: