If you pour 0.0100 kg of 20.0ºC water onto a 1.20-kg block of ice
(which is initially at −15.0ºC ), what is the final temperature? You may
assume that the water cools so rapidly that effects of the surroundings
are negligible.
To find the final temperature, we can use the principle of conservation of energy. The heat lost by the water is equal to the heat gained by the ice.
The heat lost by the water can be calculated using the equation:
Q = mcΔT
Where Q is the heat lost, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
m_water = 0.0100 kg
c_water = 4200 J/kg°C (specific heat capacity of water)
ΔT_water = (20.0°C - Tf) (since the water cools down)
The heat gained by the ice can be calculated using the equation:
Q = mcΔT
Where Q is the heat gained, m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature.
Given:
m_ice = 1.20 kg
c_ice = 2100 J/kg°C (specific heat capacity of ice)
ΔT_ice = -15.0°C - Tf (since the ice warms up)
Since the heat lost by the water is equal to the heat gained by the ice, we can set up the equation:
m_water * c_water * ΔT_water = m_ice * c_ice * ΔT_ice
Substituting the given values, we get:
0.0100 kg * 4200 J/kg°C * (20.0°C - Tf) = 1.20 kg * 2100 J/kg°C * (-15.0°C - Tf)
Simplifying the equation:
42 J/°C * (20.0°C - Tf) = -31.5 J/°C * (-15.0°C - Tf)
840°C - 42°C * Tf = 472.5°C + 31.5°C * Tf
73.5°C * Tf = 331.5°C
Tf = 4.507...
Therefore, the final temperature is approximately 4.51°C.