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a metal block of density 6000kg/m^3 n mass 1.2kg is suspended thru a spring
of spring constant 200N/m . d spring bloc system is dipped in water kept in a vessel . d water has a mass of 260g n d block is at a height 40cm above d bottom of d vessel . if d support to d spring is broken
what will b d rise in temp of d water
specific capacity of d block is 250J/kg-K n that of d water is 4200J/kg-K
heat capacities of d spring n d vessel r negligible

Amit Saxena , 12 Years ago
Grade upto college level
anser 1 Answers
Askiitians Tutor Team

To solve the problem of how much the temperature of the water will rise when the support to the spring is broken, we need to analyze the energy transformations involved. The metal block, when released, will fall and convert its potential energy into heat energy, which will then be transferred to the water. Let's break this down step by step.

Understanding the System

We have a metal block with the following properties:

  • Density: 6000 kg/m³
  • Mass: 1.2 kg
  • Specific heat capacity: 250 J/kg-K

The block is suspended by a spring with a spring constant of 200 N/m, and it is positioned 40 cm above the bottom of a vessel containing 260 g of water. The specific heat capacity of water is 4200 J/kg-K.

Calculating Potential Energy

When the support to the spring is broken, the block will fall a distance of 0.4 m (40 cm). The potential energy (PE) of the block at this height can be calculated using the formula:

PE = mgh

  • m: mass of the block = 1.2 kg
  • g: acceleration due to gravity ≈ 9.81 m/s²
  • h: height = 0.4 m

Substituting the values:

PE = 1.2 kg × 9.81 m/s² × 0.4 m = 4.7052 J

Energy Transfer to Water

This potential energy will be converted into heat energy, which will be absorbed by the water. The mass of the water is 260 g, which is equivalent to 0.26 kg. We can use the formula for heat transfer:

Q = mcΔT

Where:

  • Q: heat energy transferred (in joules)
  • m: mass of the water = 0.26 kg
  • c: specific heat capacity of water = 4200 J/kg-K
  • ΔT: change in temperature (in Kelvin or Celsius)

Setting Up the Equation

Since the potential energy of the block is converted into heat energy for the water, we can set the potential energy equal to the heat energy absorbed by the water:

4.7052 J = 0.26 kg × 4200 J/kg-K × ΔT

Solving for Temperature Change

Now, we can solve for ΔT:

ΔT = 4.7052 J / (0.26 kg × 4200 J/kg-K)

ΔT = 4.7052 J / 1092 J/K ≈ 0.0043 K

Final Result

The rise in temperature of the water when the support to the spring is broken is approximately 0.0043 K, or 4.3 mK. This indicates a very slight increase in temperature, which is typical given the relatively small mass of the block and the large heat capacity of the water.

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