Chemistry, 13.04.2020 20:38 salam6809

A 90.0 g piece of metal, initially at 98.6°C, is placed into 120.0 g of
water initially at 24.3°C. If the final temperature of the water is 34.0°C,
what is the specific heat of the metal? (The specific heat of water is 4.18
J/g.°C).

Answers: 2

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Answer from: stephaniero6

The specific heat capacity of the metal is 0.837 J/g⁰C.

The given parameters;

mass of the metal, m = 90 ginitial temperature of the metal, t₁ = 98.6⁰Cinitial temperature of the water, t₂ = 24.3°Cfinal temperature of the mixture, t = 34 ⁰Cspecific heat capacity of water, Cw = 4.18 J/g⁰C

The specific heat capacity of the metal is calculated as follows;

Heat lost by the metal = Heat gained by the water

$m_m\times c_m \times (t_1 - t) = m_w \times c_w \times (t- t_2)\\\\90\times c_m \times (98.6 - 34) = 120 \times 4.18\times (34-24.3)\\\\5814c_m = 4865.52\\\\c_m = \frac{4865.52}{5814} \\\\c_m = 0.837 \ J/g^0C$

Thus, the specific heat capacity of the metal is 0.837 J/g⁰C.

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Answer from: wellllm

specific heat of metal piece is 0.839 J/g. C .

Explanation:

As it is problem of convective heat and mass transfer, concept of energy balance can be used. The phenomenon occur here is natural. As temperature of metal is higher than that of water, so flow of temperature is from higher temperature to lower temperature, heat will flow from metal piece to water.

Data Given are as follows:

mass of piece of metal = 90 g

mass of water = 120 g

specific heat of water = 4.18 J/g.°C

specific heat of metal = ?

T metal,i = 98.6°C ...Initial temperature of metal piece

T water,i = 24.3°C ...Initial temperature of water

T metal ,f = 34°C ...Initial temperature of metal piece

T water,f = 34°C ...Initial temperature of water

ΔT water = T water,f - T water,i = 34°C - 24.3°C

ΔT metal = T metal,i - T metal,f = 98.6°C - 34°C

It is easily understood that whatever amount of heat is liberated by metal piece absorbed by water. Heat transfer is limited to thermal equilibrium between water and metal piece. Final condition is such that temperature of water will rise upto 34°C and temperature of metal will drop to 34°C.

Heat lost by metal = Heat gained by water

$m \times cm \times T metal = m \times cw \times T water$

Putting values in above equation,

$90 \times cm \times 64.4= 120 \times 4.18 \times 9.7$

$90 \times cm \times 64.4= 4865.52 \\ 5796 \times cm = 4865.52 \\ cm= \frac{4865.52}{5796} \\ cm= 0.839 J/g. C$

Therefore, specific heat of metal piece is 0.839 J/g. C .

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A 90.0 g piece of metal, initially at 98.6°C, is placed into 120.0 g of
water initially at 24....