A 63.8 kg block of silver is at initially at 107 C. 1.0 kg of ice at its freezing point is placed onto the block to cool it. As the ice melts, the water is free to flow off (so that the water never increases in temperature). After all the ice melts, what is the final temperature of the silver? Round answers to the nearest whole number, and in degrees Celsius.
use:
ΣmcΔT=0
Silver: Ti=107, Tf=T, m=63.8kg, c=240 J/kg/°C
Ice: mass=1 kg, Heat of fusion: 334000 J/kg (ΔT=0)
63.8*(240)*ΔT + 1*334000=0
ΔT = -334000/(63.8*240)=-21.8°C
Calculate Tf=Ti+ΔT
To find the final temperature of the silver, we need to use the principle of conservation of energy. The heat lost by the silver block will be equal to the heat gained by the ice that melts.
First, let's determine the heat lost by the silver block. We can use the formula: Q = m * c * ΔT, where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The specific heat capacity of silver is 0.24 J/g°C (or 240 J/kg°C). So, the heat lost by the silver block can be calculated as follows:
Q_silver = m_silver * c_silver * ΔT_silver
= 63.8 kg * 240 J/kg°C * (T_silver - 107°C)
Second, let's determine the heat gained by the ice. The heat gained by the ice is equivalent to the heat lost by the silver and can be calculated using the heat of fusion formula: Q = m * L_f, where L_f is the heat of fusion.
The heat of fusion of ice is 333.55 J/g (or 333,550 J/kg). So, the heat gained by the ice can be calculated as follows:
Q_ice = m_ice * L_f
= 1.0 kg * 333,550 J/kg
Since the water is free to flow off as it melts, the temperature of the water remains constant at the freezing point (0°C).
Now, equating the heat lost by the silver to the heat gained by the ice, we have:
Q_silver = Q_ice
63.8 kg * 240 J/kg°C * (T_silver - 107°C) = 1.0 kg * 333,550 J/kg
Let's solve this equation to find the final temperature of the silver (T_silver).
63.8 kg * 240 J/kg°C * (T_silver - 107°C) = 1.0 kg * 333,550 J/kg
(T_silver - 107°C) = (1.0 kg * 333,550 J/kg) / (63.8 kg * 240 J/kg°C)
T_silver - 107°C = 5.5°C (approximately)
T_silver ≈ 112°C
Therefore, the final temperature of the silver block is approximately 112°C.