Answer to Question #140927 in General Chemistry for ady

Question

A 125-g sample of cold water is mixed with an unknown amount of hot water in an insulated thermos bottle and allowed to equilibrate. The initial temperature of the cold and hot water are 3.0 °C and 91.0 °C, respectively. They reach an equilibrium temperature of 64.0 °C. Find the mass of hot water mixed with the cold water.

Answer 1

Q140927

A 125-g sample of cold water is mixed with an unknown amount of hot water in an insulated thermos bottle and allowed to equilibrate. The initial temperature of the cold and hot water are 3.0 °C and 91.0 °C, respectively. They reach an equilibrium temperature of 64.0 °C. Find the mass of hot water mixed with the cold water.

Solution :

We are not given any information about the thermos bottle, so we can consider that the heat exchange between cold and hot water is fast compared to the heat transfer to the thermos bottle.

When we mix the cold and hot water, cold water will gain the heat and hot water will loose some heat till an equilibrium temperature is achieved.

Cold water will gain heat from hot water, so

Heat gained by cold water = Heat loss by hot water. ---------------- Equation 1

Heat gained by cold water = m_cold * s * ΔT_{cold water} = m_cold * s * (T _Final - T_cold )

Heat loss by hot water = m _Hot * s * ΔT_{hot water} = m _Hot * s * ( T_hot - T _Final )

where ‘s’ is the specific heat capacity of water.

Substitute this in Equation 1, we have

m_cold * s * (T _Final - T_cold ) = m _Hot * s * ( T_hot - T _Final ) --------------------- Equation 2

We know,

mass of cold water, m_cold = 125g

Temperature of cold water, T_cold = 3.0⁰C

specific heat capacity of water, s = 4.184 J/g ⁰C.

Temperature of Hot water, T_Hot = 91.0⁰C

Equilibrium temperature, T _Final = 64.0 ⁰C

Plug all this information in Equation 2, we have

125g * 4.184 J/g ⁰C * ( 64.0 ⁰C - 3.0⁰C) = m _Hot * 4.184 J/g ⁰C * (91.0⁰C-64.0 ⁰C);

125g * 4.184 J/g ⁰C * 61.0⁰C = m _Hot * 4.184 J/g ⁰C * 27.0⁰C

31903J = m _Hot * 112.968J/g

divide both the side by 112.968J/g, we have

31903J/ 112.968J/g = m _Hot * 112.968J/g / 112.968J/g

282.407g = m _Hot ;

The quantities given in the question are in 3 significant figures, so our final answer must also be in 3 significant figures.

Hence, mass of hot water, m _Hot = 282 g ;