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Grade 11Thermal Physics

A block of density 6000kg/m^3 and mass 1.2 kg is suspended through a spring of spring constant 200N/m . The spring block system is dipped in water kept in a vessel . The water has a mass of 260 g and the block is at the height of 40 cm above the bottom of the vessel . If the support to the spring is broken , what will be the rise in temperature of the water . Specific heat capacity of block is 250 J/kg-K and that of water is 4200 J/kg-K . Heat capacity of spring and vessel is negligible.

Profile image of Harsh Kakran
8 Years agoGrade 11
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1 Answer

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ApprovedApproved Tutor Answer1 Year ago

To determine the rise in temperature of the water when the support to the spring is broken, we need to analyze the energy transfer that occurs when the block falls into the water. The potential energy of the block will be converted into heat energy, which will then raise the temperature of the water. Let's break this down step by step.

Step 1: Calculate the Potential Energy of the Block

The potential energy (PE) of the block when it is suspended at a 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 above the water = 0.40 m

Substituting the values:

PE = 1.2 kg × 9.81 m/s² × 0.40 m = 4.7052 J

Step 2: Determine the Heat Transfer to the Water

When the block falls into the water, the potential energy will be converted into heat energy, which will be absorbed by the water and the block. However, since the block is initially at a different temperature than the water, we need to consider the specific heat capacities to find the temperature change.

Heat Transfer Equation

The heat gained by the water can be expressed as:

Q = mcΔT

  • m = mass of the water = 0.260 kg
  • c = specific heat capacity of water = 4200 J/kg-K
  • ΔT = change in temperature of the water

Step 3: Calculate the Temperature Rise

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

PE = Q

Thus, we have:

4.7052 J = 0.260 kg × 4200 J/kg-K × ΔT

Now, we can solve for ΔT:

ΔT = 4.7052 J / (0.260 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 about 4.3 mK. This indicates a very slight increase in temperature, which is typical given the relatively small amount of energy transferred compared to the mass and heat capacity of the water.