How many joules would be required to raise the temperature of 20g of ice at oC to water at15C?
Q1= rm=330000 x 0.02 = …
Q2=cm(T2-T1)=4180 x 0.02 x 15 = ...
Q= Q1 +Q2 = …
To answer this question, you need to calculate the total heat energy required to raise the temperature of the ice from 0°C to water at 0°C and then further raise it to 15°C.
To calculate the heat energy, you can use the specific heat capacity (C) of ice and water, along with the formula:
Q = m × C × ΔT
where:
Q is the heat energy (in joules),
m is the mass (in grams),
C is the specific heat capacity (in J/g°C), and
ΔT is the change in temperature (in °C).
First, calculate the heat energy required to raise the temperature of the ice from 0°C to 0°C (melting point of ice):
Q1 = m × Cice × ΔT1
Since the temperature is constant, ΔT1 = 0, so the heat energy required to melt the ice is 0.
Next, calculate the heat energy required to raise the temperature of the ice (now water) from 0°C to 15°C:
Q2 = m × Cwater × ΔT2
Given:
m (mass of ice) = 20g
Cice (specific heat capacity of ice) = 2.09 J/g°C
Cwater (specific heat capacity of water) = 4.18 J/g°C
ΔT2 (change in temperature) = 15°C
Substituting the values into the equation:
Q2 = 20g × 4.18 J/g°C × 15°C
Q2 ≈ 1254 joules
Therefore, it would require approximately 1254 joules of heat energy to raise the temperature of 20g of ice at 0°C to water at 15°C.