# An object with a mass of #12 kg#, temperature of #210 ^oC#, and a specific heat of #7 J/(kg*K)# is dropped into a container with #32 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

#40572-84T=36.51648+0.13376T

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To determine if the water evaporates, we need to calculate the amount of heat transferred from the object to the water and compare it to the heat required for the water to evaporate.

First, calculate the heat transferred from the object to the water using the formula:

[ Q = mc\Delta T ]

Where:

- ( Q ) is the heat transferred,
- ( m ) is the mass of the object,
- ( c ) is the specific heat capacity,
- ( \Delta T ) is the change in temperature.

Given:

- ( m_{\text{object}} = 12 , \text{kg} ),
- ( T_{\text{initial}} = 210 , ^\circ C ),
- ( c_{\text{object}} = 7 , \text{J/(kg*K)} ),
- ( T_{\text{water, initial}} = 0 , ^\circ C ),
- ( V_{\text{water}} = 32 , \text{L} ).

First, we need to convert the volume of water to mass using the density of water:

[ \rho_{\text{water}} = 1000 , \text{kg/m}^3 ] [ V_{\text{water}} = 32 , \text{L} = 0.032 , \text{m}^3 ] [ m_{\text{water}} = \rho_{\text{water}} \times V_{\text{water}} = 1000 \times 0.032 = 32 , \text{kg} ]

Now, calculate the heat transferred:

[ Q = m_{\text{object}} \times c_{\text{object}} \times (T_{\text{final}} - T_{\text{initial}}) ] [ Q = 12 \times 7 \times (0 - 210) = -17640 , \text{J} ]

Since the heat transferred is negative, it indicates that heat is transferred from the object to the water.

Next, we need to calculate the heat required to evaporate the water:

[ Q_{\text{evaporation}} = m_{\text{water}} \times L_{\text{vaporization}} ]

The latent heat of vaporization for water is ( L_{\text{vaporization}} = 2260 , \text{J/kg} ).

[ Q_{\text{evaporation}} = 32 \times 2260 = 72320 , \text{J} ]

Comparing the heat transferred from the object (-17640 J) to the heat required to evaporate the water (72320 J), we find that the heat transferred is insufficient to evaporate the water. Therefore, the water does not evaporate.

To find out by how much the water's temperature changes, we can use the fact that the heat gained by the water is equal to the heat lost by the object:

[ Q_{\text{water}} = -Q_{\text{object}} ] [ m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}} = -m_{\text{object}} \times c_{\text{object}} \times (T_{\text{final}} - T_{\text{initial}}) ]

Solving for ( \Delta T_{\text{water}} ):

[ \Delta T_{\text{water}} = \frac{-m_{\text{object}} \times c_{\text{object}} \times (T_{\text{final}} - T_{\text{initial}})}{m_{\text{water}} \times c_{\text{water}}} ]

Given:

- ( c_{\text{water}} = 4186 , \text{J/(kg*K)} )

Substitute the values and solve:

[ \Delta T_{\text{water}} = \frac{-12 \times 7 \times (T_{\text{final}} - 210)}{32 \times 4186} ]

[ \Delta T_{\text{water}} = \frac{-84 \times (T_{\text{final}} - 210)}{133792} ]

[ \Delta T_{\text{water}} = \frac{-84T_{\text{final}} + 17640}{133792} ]

[ \Delta T_{\text{water}} = \frac{-84T_{\text{final}}}{133792} + \frac{17640}{133792} ]

[ \Delta T_{\text{water}} = -0.000628T_{\text{final}} + 0.1319 ]

Therefore, the water's temperature changes by ( -0.000628T_{\text{final}} + 0.1319 ) degrees Celsius.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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