## A 125.0-g sample of a metal heated to 100.0 ∘C and placed in a calorimeter that contains 250.0 g of water. The temperature rises from 24.3 ∘

Question

A 125.0-g sample of a metal heated to 100.0 ∘C and placed in a calorimeter that contains 250.0 g of water. The temperature rises from 24.3 ∘C to 27.2 ∘C. What is the specific heat capacity of the metal? Ignore the calorimeter in your analysis. Group of answer choices

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5 months 2021-08-16T05:21:10+00:00 1 Answers 5 views 0

0.333J/g°C is the specific heat of the metal

Explanation:

The heat that the metaal gives is equal to the heat that water is absorbing. The equation is:

S(metal)*ΔT(metal)*Mass(metal) = S(H2O)*ΔT(H2O)*Mass(H2O)

Where S is specific heat, ΔT is change in temperature and mass the mass in grams of the metal and water.

Replacing:

S(metal)*(100.0°C-27.2°C)*125.0g = 4.184J/g°C*(27.2°C-24.3°C)*250.0g

S(metal) = 4.184J/g°C*(27.2°C-24.3°C)*250.0g / (100.0°C-27.2°C)*125.0g

S(metal) = 0.333J/g°C is the specific heat of the metal