matty374
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A 50 g sample of an unknown metal is heated to 90.0C. It is placed in a perfectly insulated container along with 100 g of water at an initial temperature of 20C. After a short time, the temperature of both the metal and water become equal at 25C. The specific heat of water is 4.18 J/gC in this temperature range. What is the specific heat capacity of the metal?
Record your answer with two significant figures. J/gC

Respuesta :

Q = mcΔT
Qwater = -Qmetal

(100 g)(4.18 J/g°C)(25°C - 20°C) = -(50 g)c(25°C - 90.0°C)

2090 J = (3250 g °C)c

c = 0.643 J/g °C
Eduard22sly

The specific heat capacity of the metal, given the data is 0.6 J/gºC

Data obtained From the question

  • Mass of metal (M) = 50 g
  • Temperature of metal (T) = 90 °C
  • Mass of water (Mᵥᵥ) = 100 g
  • Temperature of water (Tᵥᵥ) = 20 °C
  • Equilibrium temperature (Tₑ) = 25 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.18 J/gºC
  • Specific heat capacity of gold (C) =?

How to determine the specific heat capacity of the metal

Heat loss = Heat gain

MC(T –Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)

50 × C × (90 – 25) = 100 × 4.18 × (25 – 20)

C × 3250 = 2090

Divide both side by 3250

C = 2090 / 3250

C = 0.6 J/gºC

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