a piece of silver metal is taken from a furnace where the temperature is 75C and placed on large block of ice T 0.0C. assuming that all heat is given up by the silver is used to melt the ice,how much ice is melted?
Csilver * mass of silver * 75 = mass of ice * heat of fusion of water
Summation of Q = 0
Qsilver + Qice = 0
Qsilver = (mass of silver)(Csilver)(final temp - initial temp)
Use Latent heat because ice transformed phase to water
Qice = (mass of ice)(heat fusion of water)
mass of silver * Csilver * (0-75) + mass of ice * heat fusion of water
Isolate and you'll get:
mass of ice * heat fusion of water = Csilver * mass of silver * 75
mass of ice = (Csilver * mass of silver * 75)/heat fusion of water
To determine how much ice is melted when a piece of silver metal is placed on a large block of ice, you can use the concept of heat transfer and the heat equation. The heat equation is given by:
Q = m * c * ΔT
Where:
Q is the heat transfer
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
In this case, the heat transfer (Q) from the silver metal will be equal to the heat required to melt the ice.
First, you need to determine the heat transfer for the silver metal. The specific heat capacity (c) of silver is 0.235 J/g°C. Let's assume the mass of the silver metal is 100 grams.
Q_silver = m_silver * c_silver * ΔT_silver
= 100 g * 0.235 J/g°C * (75°C - 0°C)
= 1762.5 J
Now, you need to determine the heat required to melt the ice. The heat of fusion (or latent heat) for ice is 334 J/g. This means that it takes 334 J of heat to convert 1 gram of ice at 0°C into water at 0°C.
Q_ice = m_ice * ΔH_fusion
Since all the heat is used to melt the ice, we can set Q_silver equal to Q_ice:
Q_silver = Q_ice
m_ice * ΔH_fusion = 1762.5 J
Solving for the mass of ice (m_ice):
m_ice = 1762.5 J / ΔH_fusion
= 1762.5 J / 334 J/g
≈ 5.27 g
Therefore, approximately 5.27 grams of ice will be melted when the piece of silver metal is placed on the large block of ice.