Question

A 4.82g of an unknown metal is heated to 115°C and then placed in 35mL of water at 28.7°C, which then heats up to 34.5°C. What is the specific heat of the metal? Ans. 21888.99 J/kg- °C 2189J / kg ° C

Answer 1

To find the specific heat of the unknown metal, we can use the equation for heat transfer:

Q = m * c * ΔT

Where: Q is the heat transferred m is the mass of the metal c is the specific heat of the metal ΔT is the change in temperature

We can break down the problem into two parts: the heat absorbed by the water and the heat released by the metal.

1. Heat absorbed by the water: Q_water = m_water * c_water * ΔT_water

Given values for water: m_water = 35 g (since the volume is given as 35 mL, and assuming the density of water as 1 g/mL) c_water = 4.18 J/g°C (specific heat capacity of water) ΔT_water = (34.5°C - 28.7°C) = 5.8°C

Q_water = 35 g * 4.18 J/g°C * 5.8°C = 853.97 J

2. Heat released by the metal: Q_metal = m_metal * c_metal * ΔT_metal

Given values for the metal: m_metal = 4.82 g ΔT_metal = (115°C - 34.5°C) = 80.5°C

We want to find c_metal, so we rearrange the equation to solve for it:

c_metal = Q_metal / (m_metal * ΔT_metal)

c_metal = 853.97 J / (4.82 g * 80.5°C) ≈ 21888.99 J/kg-°C (rounded to five significant figures)

Therefore, the specific heat of the unknown metal is approximately 21888.99 J/kg-°C or approximately 2189 J/kg°C (rounded to three significant figures).