Question: In a calorimetry experiment, if 100 g of water absorbs 4200 J of heat, what is the change in temperature assuming no heat loss?
Options:
1 °C
2 °C
3 °C
4 °C
Correct Answer: 1 °C
Solution:
Using q = mcΔT, ΔT = q/(mc) = 4200 J / (100 g * 4.18 J/g°C) = 10 °C.
In a calorimetry experiment, if 100 g of water absorbs 4200 J of heat, what is t
Practice Questions
Q1
In a calorimetry experiment, if 100 g of water absorbs 4200 J of heat, what is the change in temperature assuming no heat loss?
1 °C
2 °C
3 °C
4 °C
Questions & Step-by-Step Solutions
In a calorimetry experiment, if 100 g of water absorbs 4200 J of heat, what is the change in temperature assuming no heat loss?
Step 1: Identify the formula to use, which is q = mcΔT, where q is the heat absorbed, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature.
Step 2: Write down the values you have: q = 4200 J (heat absorbed), m = 100 g (mass of water), and c = 4.18 J/g°C (specific heat capacity of water).
Step 3: Rearrange the formula to solve for ΔT: ΔT = q / (mc).
Step 4: Substitute the values into the rearranged formula: ΔT = 4200 J / (100 g * 4.18 J/g°C).
Step 5: Calculate the denominator: 100 g * 4.18 J/g°C = 418 J/°C.
Step 6: Now divide the heat absorbed by the result from Step 5: ΔT = 4200 J / 418 J/°C.
Step 7: Perform the division: ΔT = 10 °C.
Calorimetry – The study of heat transfer during physical and chemical processes, often using the formula q = mcΔT, where q is heat absorbed or released, m is mass, c is specific heat capacity, and ΔT is the change in temperature.
Specific Heat Capacity – The amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius.
Heat Transfer – The process of thermal energy moving from a hotter object to a cooler one, which is assumed to be negligible in this problem.
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