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What is the temperature change of 2.00-kg of water if 505-J if thermal energy is added to it?

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Final answer:

The temperature change of 2.00-kg water when 505 J of thermal energy is added to it is approximately 0.0604 °C.

Step-by-step explanation:

The temperature change of 2.00 kg of water can be calculated using the specific heat capacity formula:

ΔQ = mcΔT

Where:

  • ΔQ is the change in thermal energy, which is given as 505 J
  • m is the mass of the water, which is 2.00 kg
  • c is the specific heat capacity of water, which is 4.184 J/g °C or 4184 J/kg °C
  • ΔT is the change in temperature, which we need to calculate

Substituting the given values into the formula:

505 J = (2.00 kg)(4184 J/kg °C)(ΔT)

Solving for ΔT:

ΔT = 505 J / (2.00 kg * 4184 J/kg °C)

ΔT = 0.0604 °C

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