When 47.5J of heat are added to 13.2 g of liquid, its temperature rises by 1.72 degrees C. What is the heat capacity of the liquid?
A cook wants to heat 1.35kg of water from 32.0 degrees C to 100 degrees C. If he uses combustion of natural gas to heat the water, hou much natural gas will he need to burn? Natural gas produces 49.3kJ of heat per gram. (for the sake of simplicity assume that the transger of heat is 100% efficient.)
Thanks:)
1.
q = mass liquid x sp. h. x (Tfinal-Tinitial)
Solve for sp.h.
2.
How much heat is needed? That is
q = mass x specific heat x (Tfinal-Tinitial). Solve for q.
Then CH4 + 2O2 ==> CO2 + 2H2O
16 g CH4 x (q in Joules/49,300 J) = ? g CH4.
To find the heat capacity of the liquid in the first question, we need to use the formula:
Heat capacity = Q / (m * ΔT)
Where:
Q = heat added to the liquid (47.5 J)
m = mass of the liquid (13.2 g)
ΔT = change in temperature (1.72 °C)
Substituting the values:
Heat capacity = 47.5 J / (13.2 g * 1.72 °C)
Calculating the value:
Heat capacity = 47.5 J / 22.704 g°C
Heat capacity = 2.09 J/g°C
Therefore, the heat capacity of the liquid is 2.09 J/g°C.
For the second question, we need to calculate the amount of heat required to heat the water using the formula:
Q = m * C * ΔT
Where:
Q = heat required
m = mass of the water (1.35 kg)
C = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (100 - 32 = 68 °C)
Substituting the values:
Q = 1.35 kg * 1000 g/kg * 4.18 J/g°C * 68 °C
Calculating the value:
Q = 377,940 J
Since we are given that the transfer of heat is 100% efficient, all the heat produced by burning natural gas will be used to heat the water. So, we need to find how many grams of natural gas will produce 377,940 J of heat.
To do that, we divide the heat required by the heat produced per gram of natural gas:
Number of grams of natural gas = Q / heat produced per gram
Number of grams of natural gas = 377,940 J / 49.3 kJ/g (note that 1 kJ = 1000 J)
Calculating the value:
Number of grams of natural gas = 7.67 g
Therefore, the cook will need to burn approximately 7.67 grams of natural gas to heat the water from 32.0°C to 100°C.
To find the heat capacity of the liquid, we can use the formula:
Heat Capacity = Heat Energy / Change in Temperature
Given:
Heat Energy = 47.5 J
Change in Temperature = 1.72 °C
First, we need to convert the mass of the liquid to grams:
13.2 g
Now, we can calculate the heat capacity:
Heat Capacity = 47.5 J / 1.72 °C
Heat Capacity = 27.616 J/°C (rounded to three decimal places)
Therefore, the heat capacity of the liquid is 27.616 J/°C.
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To calculate the amount of natural gas the cook needs to burn, we can use the equation:
Heat Energy = Mass of Natural Gas Burned * Heat Produced per gram
Given:
Heat Energy = Unknown
Mass of Water = 1.35 kg
Initial Temperature = 32.0 °C
Final Temperature = 100 °C
Heat Produced per gram of Natural Gas = 49.3 kJ/g
First, we need to convert the mass of water to grams:
1.35 kg = 1350 g
Next, we calculate the heat energy:
Heat Energy = Mass of Water * Specific Heat Capacity * Change in Temperature
Specific Heat Capacity of Water = 4.186 J/g°C
Change in Temperature = Final Temperature - Initial Temperature
Change in Temperature = (100 °C - 32.0 °C) = 68 °C
Heat Energy = 1350 g * 4.186 J/g°C * 68 °C
Heat Energy = 377,661.84 J (rounded to two decimal places)
Now, we can calculate the mass of natural gas burned:
Mass of Natural Gas Burned = Heat Energy / Heat Produced per gram
Mass of Natural Gas Burned = 377,661.84 J / 49.3 kJ/g
Converting kJ to J:
1 kJ = 1000 J
Mass of Natural Gas Burned = 377,661.84 J / (49.3 kJ/g * 1000)
Mass of Natural Gas Burned = 7.6597 g (rounded to four decimal places)
Therefore, the cook needs to burn approximately 7.6597 grams of natural gas.