When 20 kJ of heat is removed from 1.2 kg of ice originally at -15 C, what is the new temperature?
5 answers
I know or I think the formula is Q=mcT, but not sure how to use it in this problem.
ice is about 2.03 Joules/ gram deg c
so 2030 J/kg deg c
1.2 kg *2030 J /kg degC (-15-T) = 20,000 J
so 2030 J/kg deg c
1.2 kg *2030 J /kg degC (-15-T) = 20,000 J
I am still confused on how to work this problem to get the temperature.
23•c
To solve this problem, we can use the formula for the heat transfer:
Q = mcΔT
Where:
Q = heat transferred (in J)
m = mass of the substance (in kg)
c = specific heat capacity (in J/kg°C)
ΔT = change in temperature (in °C)
In this case, the substance is ice, so we will use the specific heat capacity of ice, which is approximately 2.03 J/g°C or 2030 J/kg°C.
Given:
Q = -20 kJ = -20,000 J (since heat is being removed)
m = 1.2 kg
c = 2030 J/kg°C
ΔT = unknown
Rearranging the formula, we have:
ΔT = Q / (mc)
Substituting in the known values:
ΔT = -20,000 J / (1.2 kg * 2030 J/kg°C)
ΔT ≈ -8.21 °C
The change in temperature is -8.21 °C. To find the new temperature, we need to subtract this change from the initial temperature of -15 °C:
New temperature = -15 °C - (-8.21 °C)
New temperature ≈ -6.79 °C
Therefore, the new temperature of the ice after 20 kJ of heat is removed is approximately -6.79 °C.
Q = mcΔT
Where:
Q = heat transferred (in J)
m = mass of the substance (in kg)
c = specific heat capacity (in J/kg°C)
ΔT = change in temperature (in °C)
In this case, the substance is ice, so we will use the specific heat capacity of ice, which is approximately 2.03 J/g°C or 2030 J/kg°C.
Given:
Q = -20 kJ = -20,000 J (since heat is being removed)
m = 1.2 kg
c = 2030 J/kg°C
ΔT = unknown
Rearranging the formula, we have:
ΔT = Q / (mc)
Substituting in the known values:
ΔT = -20,000 J / (1.2 kg * 2030 J/kg°C)
ΔT ≈ -8.21 °C
The change in temperature is -8.21 °C. To find the new temperature, we need to subtract this change from the initial temperature of -15 °C:
New temperature = -15 °C - (-8.21 °C)
New temperature ≈ -6.79 °C
Therefore, the new temperature of the ice after 20 kJ of heat is removed is approximately -6.79 °C.