An object with a mass of #2 kg#, temperature of #214 ^oC#, and a specific heat of #12 (KJ)/(kg*K)# is dropped into a container with #25 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

An object with a mass of #2 kg#, temperature of #214  ^oC#, and a specific heat of #12 (KJ)/(kg*K)# is dropped into a container with #25  L # of water at #0^oC #.  Does the water evaporate?  If not, by how much does the water's temperature change?

Layla Ahmed · Accepted Answer

The water will not evaporate and the change in temperature is #=39.9^@C#

The heat is transferred from the hot object to the cold water.

Let #T=# final temperature of the object and the water

For the cold water, # Delta T_w=T-0=T#

For the object #DeltaT_o=214-T#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

The specific heat of the object is #C_o=12kJkg^-1K^-1#

The mass of the object is #m_0=2kg#

The volume of water is #V=25L#

The density of water is #rho=1kgL^-1#

The mass of the water is #m_w=rhoV=25kg#

#2*12*(214-T)=25*4.186*T#

#214-T=(25*4.186)/(2*12)*T#

#214-T=4.36T#

#5.36=214#

#T=214/5.36=39.9^@C#

As the final temperature is #T<100^@C#, the water will not evaporate.

Axel Acevedo · Answer

undefined To determine if the water evaporates, calculate the heat absorbed by the water using the formula: $ Q = mcΔT $, where $ Q $ is the heat absorbed, $ m $ is the mass, $ c $ is the specific heat, and $ ΔT $ is the change in temperature. Then compare $ Q $ with the heat required for the water to evaporate. If $ Q $ is less than the heat required for evaporation, the water does not evaporate. If $ Q $ is greater, then the water evaporates. If the water does not evaporate, use the same formula to calculate $ ΔT $.

Layla Ahmed · Answer

undefined The water will not evaporate. The temperature change of the water can be calculated using the formula:

$$ Q = mcΔT $$

Where:
- Q is the heat transferred (in joules or kilojoules)
- m is the mass of the water (in kilograms)
- c is the specific heat capacity of water (4.18 kJ/kg*K)
- ΔT is the change in temperature (in degrees Celsius)

First, convert the volume of water from liters to kilograms using the density of water (1 kg/L).

$$ \text{Mass of water} = \text{Volume of water} \times \text{Density of water} $$
$$ \text{Mass of water} = 25 \, \text{L} \times 1 \, \text{kg/L} $$
$$ \text{Mass of water} = 25 \, \text{kg} $$

Next, calculate the heat transfer using the formula:

$$ Q = mcΔT $$

$$ Q = 25 \, \text{kg} \times 4.18 \, \text{kJ/kg*K} \times (214 \, \text{°C} - 0 \, \text{°C}) $$
$$ Q = 25 \, \text{kg} \times 4.18 \, \text{kJ/kg*K} \times 214 \, \text{°C} $$
$$ Q = 22045 \, \text{kJ} $$

Since the object was dropped into the water, it transferred heat to the water, causing its temperature to rise. The water will not evaporate unless additional heat is supplied to reach its boiling point.