Question

I m² of water at T-15 °C (specific heat 4186 J/Kg C) is reversed in a copper container of 0.5 Tons exposed to direct sun and having reached a temperature of 120 ° C (specific heat 390 J/Kg C). Does the water boil up? Why?

Answer 1

This question involves the concepts of specific heat capacity, the law of conservation of energy, and temperature.

The water "will not boil up" because the equilibrium temperature is well

below the boiling point of water.

To find out if the water will boil up or not, we will have to find the equilibrium temperature of the water and container system. Applying the law of conservation of energy:

Energy Lost by Container = Energy Gained by Water

$m_cC_c\Delta T_c=m_wC_w\Delta T_w$

where,

$m_c$ = mass of container = 0.5 ton = 453.6 kg

$m_w$ = mass of water = (density)(volume) = (1000 kg/m³)(1 m³) = 1000 kg

$C_c$ = specific heat capacity of copper = 390 J/kg.°C

$C_w$ = specific heat capacity of water = 4186 J/kg.°C

$\Delta T_c$ = change in temperature of copper container = 120°C - T

$\Delta T_w$ = change in temperature of water = T - 15°C

T = Equilibrium temperature = ?

Therefore,

$(453.6\ kg)(390\ J/kg.^oC)(120^oC-T)=(1000\ kg)(4186\ J/kg.^oC)(T-15^oC)\\\\4186000T\ J/^oC + 176904T\ J/^oC=21228480\ J+62790000\ J\\\\T=\frac{84018480\ J}{4362904\ J/^oC}$

T = 19.25°C

Since this equilibrium temperature is well below the boiling point of water. Hence, the water will not boil.

Learn more about specific heat capacity here:

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