Question

You are measuring the specific heat of a material T. You heat a 215 g sample of the metal to 100°C and place the sample in 450 g of water at 20°C. The material and water reach an equilibrium temperature of 21.5°C. What is the specific heat of the material T? CpWater = 4.18 J/ (g*C)

Answer 1

The specific heat of the material is: 0.49 J/g°C

What is the specific heat of the material?

To find the specific heat of the material T, we can use the formula:

q = m × c × ΔT

where:

(q) = heat gained or lost

(m) = mass

(c) = specific heat

(ΔT) = change in temperature

First, we need to find the heat lost by the metal and gained by the water. The heat lost by the metal is equal to the heat gained by the water, so we have:

water

m_metal × c_metal × ΔT_metal = m_water ×c_water ×ΔT_water

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Substituting the given values, we can solve for
`c_{\text{metal}}`, the specific heat of the metal.

Given:

`m_{\text{metal}} = 215 , \text{g}`

`c_{\text{water}} = 4.18 , \text{J/g°C}`

`m_{\text{water}} = 450 \text{g}`

`\Delta T_{\text{water}`= 1.5°C (final temperature - initial temperature)

Substituting the values:

215 × c_metal × (100 - 21.5) × (100 − 21.5) = 450 × 4.18 × 1.5

Solving for
`c_{\text{metal}`}), we get:

c_metal = 0.49 J/g°C