1). A hot iron horseshoe (mass=.40kg), just forged, is dropped into 1.35L of water in a .30kg iron pot initially at 20.0 degrees Celsius. If the final equilibrium temperature is 25.0 degrees Celsius, estimate the initial temperature of the hot horseshoe.
2). A cube of ice is taken from the freezer at -8.5 degrees Celsius and placed in a 95kg aluminum calorimeter filled with 310 g of water at room temperature of 20.0 degrees Celsius. The final situation, is observed to be all water at 17.0 degrees Celsius. What was the mass of the ice cube?
The sum of the heats gained is zero.
Heat gained by water+heatgainebysteel+ heat gained by iron pot = 0
1.35kg*cwater*(25-20)+.40(csteel)(25-Ti)+.30*ciron(25-20)=0
solve for Ti if the steel horseshoe.
To solve both of these questions, we can use the principle of conservation of energy, specifically the equation:
Amount of heat gained by the water = Amount of heat lost by the horseshoe or ice cube
1) Let's solve the first question:
The amount of heat gained by the water is given by the equation:
Q_water = m_water * C_water * ΔT_water
Where:
m_water is the mass of the water (1.35 kg)
C_water is the specific heat capacity of water (4.186 J/g°C or 4190 J/kg°C)
ΔT_water is the change in temperature of the water (final temperature - initial temperature)
The amount of heat lost by the horseshoe is given by the equation:
Q_horseshoe = m_horseshoe * C_horseshoe * ΔT_horseshoe
Where:
m_horseshoe is the mass of the horseshoe (0.40 kg)
C_horseshoe is the specific heat capacity of iron (0.449 J/g°C or 449 J/kg°C)
ΔT_horseshoe is the change in temperature of the horseshoe (final temperature - initial temperature)
At equilibrium, the amount of heat gained by the water is equal to the amount of heat lost by the horseshoe:
m_water * C_water * ΔT_water = m_horseshoe * C_horseshoe * ΔT_horseshoe
Substituting the given values, we have:
(1.35 kg) * (4190 J/kg°C) * (25.0°C - T_initial) = (0.40 kg) * (449 J/kg°C) * (T_initial - 20.0°C)
Simplifying this equation will give us the initial temperature of the horseshoe, T_initial.
2) Similarly, let's solve the second question:
The amount of heat gained by the water is given by the equation:
Q_water = m_water * C_water * ΔT_water
Where:
m_water is the mass of the water (310 g or 0.310 kg)
C_water is the specific heat capacity of water (4.186 J/g°C or 4190 J/kg°C)
ΔT_water is the change in temperature of the water (final temperature - initial temperature)
The amount of heat lost by the ice cube is given by the equation:
Q_ice = m_ice * C_ice * ΔT_ice
Where:
m_ice is the mass of the ice cube (unknown, we'll solve for it)
C_ice is the specific heat capacity of ice (2.093 J/g°C or 2093 J/kg°C)
ΔT_ice is the change in temperature of the ice cube (final temperature - initial temperature)
At equilibrium, the amount of heat gained by the water is equal to the amount of heat lost by the ice cube:
m_water * C_water * ΔT_water = m_ice * C_ice * ΔT_ice
Substituting the given values, we have:
(0.310 kg) * (4190 J/kg°C) * (17.0°C - 20.0°C) = m_ice * (2093 J/kg°C) * (20.0°C - (-8.5°C))
Simplifying this equation will give us the mass of the ice cube, m_ice.