Question

A 30.0-g piece of metal at 203oC is dropped into 400.0 g of water at 25.0 oC. The water temperature rises to 29.0 oC. Calculate the specific heat of the metal (J/goC). Assume that all of the heat lost by the metal is transferred to the water and no heat is lost to the surroundings.

Answer 1

The specific heat of the metal is 2.4733 J/g°C.

Given the following data:

• Initial temperature of water = 25.0°C

• Final temperature of water = 29.0°C

• Mass of water = 400.0 g

• Mass of metal = 30.0 g

• Temperature of metal = 203.0°C

We know that the specific heat capacity of water is 4.184 J/g°C.

To find the specific heat of the metal (J/g°C):

Heat lost by metal = Heat gained by water.

$Q_{metal} = Q_{water}$

Mathematically, heat capacity or quantity of heat is given by the formula;

$Q = mc\theta$

Where:

• Q is the heat capacity or quantity of heat.

• m is the mass of an object.

• c represents the specific heat capacity.

• ∅ represents the change in temperature.

Substituting the values into the formula, we have:

$30(203)c = 400(4.184)(29 \; -\; 20)\\\\6090c = 1673.6(9)\\\\6090c = 15062.4\\\\c = \frac{15062.4}{6090}$

Specific heat capacity of metal, c = 2.4733 J/g°C

Therefore, the specific heat of the metal is 2.4733 J/g°C.

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