Activity 2: Calculate for the unknown on the following problems
1. The temperature of 335 g of water changed from 24.5°C to 26.4°C. How much heat did this sample absorb?
2 How much heat in kilojoules has to be removed from 225g of water to lower its temperature from 25.0°C to 10.0°C?
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Answer:
1.C_water = 4.184 J/g degree C (ans. 2.66 kJ) How much heat in kilojoules has to be removed from 225g of water to lower its temperature from 25.0 degree C to 10.0 degree C
2.14108 J
Explanation for number 2:
The heat that has to be removed from the water is given by
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔT
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwhere
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwherem = 225 g is the water mass
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwherem = 225 g is the water massC_s = 4.18 J/gCC
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwherem = 225 g is the water massC_s = 4.18 J/gCC s
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwherem = 225 g is the water massC_s = 4.18 J/gCC s
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwherem = 225 g is the water massC_s = 4.18 J/gCC s =4.18J/gC is the specific heat capacity of the water
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwherem = 225 g is the water massC_s = 4.18 J/gCC s =4.18J/gC is the specific heat capacity of the water\Delta T=10.0 C-25.0 C=-15.0 CΔT=10.0C−25.0C=−15.0C is the charge in temperature of the water
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwherem = 225 g is the water massC_s = 4.18 J/gCC s =4.18J/gC is the specific heat capacity of the water\Delta T=10.0 C-25.0 C=-15.0 CΔT=10.0C−25.0C=−15.0C is the charge in temperature of the waterSubstituting into the equation, we find
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwherem = 225 g is the water massC_s = 4.18 J/gCC s =4.18J/gC is the specific heat capacity of the water\Delta T=10.0 C-25.0 C=-15.0 CΔT=10.0C−25.0C=−15.0C is the charge in temperature of the waterSubstituting into the equation, we findQ=(225 g) (4.18 J/gC) (-15.0 C)=-14,108 JQ=(225g)(4.18J/gC)(−15.0C)=−14,108J
The heat that has to be removed from the water is given byQ=m C_s \Delta TQ=mC s ΔTwherem = 225 g is the water massC_s = 4.18 J/gCC s =4.18J/gC is the specific heat capacity of the water\Delta T=10.0 C-25.0 C=-15.0 CΔT=10.0C−25.0C=−15.0C is the charge in temperature of the waterSubstituting into the equation, we findQ=(225 g) (4.18 J/gC) (-15.0 C)=-14,108 JQ=(225g)(4.18J/gC)(−15.0C)=−14,108Jand the negative sign means the heat is removed from the system.