A student uses a 350g block of tungsten to heat a 600ml beaker containing 450cm3 of tapwater. She adds exactly 2.5kJ of energy to the block before carefully placing it into the water.If the tap water was originally at 23oC, what would be the highest temperature that waterreaches? (assume that no heat is lost to the surroundings)
Question
A student uses a 350g block of tungsten to heat a 600ml beaker containing 450cm3 of tapwater. She adds exactly 2.5kJ of energy to the block before carefully placing it into the water.If the tap water was originally at 23oC, what would be the highest temperature that waterreaches? (assume that no heat is lost to the surroundings)
Solution
To solve this problem, we need to use the concept of conservation of energy, which states that energy cannot be created or destroyed, only transferred. In this case, the energy is being transferred from the tungsten block to the water.
First, we need to calculate how much the temperature of the tungsten block will increase when 2.5 kJ of energy is added. The specific heat capacity of tungsten is 0.134 J/g°C.
The formula to calculate the change in temperature is:
Q = mcΔT
where: Q = heat energy added (in joules) m = mass (in grams) c = specific heat capacity (in J/g°C) ΔT = change in temperature (in °C)
Rearranging the formula to solve for ΔT gives us:
ΔT = Q / (mc)
Substituting the given values:
ΔT = 2500 J / (350 g * 0.134 J/g°C) = 53.5°C
So, the tungsten block will heat up to 53.5°C.
Next, we need to calculate how much this will heat up the water. The specific heat capacity of water is 4.18 J/g°C, and 1 cm³ of water has a mass of approximately 1 g, so the mass of the water is 450 g.
Using the same formula, but solving for Q (because we want to find out how much heat energy will be transferred to the water), we get:
Q = mcΔT
Substituting the values for the water and the change in temperature of the tungsten block:
Q = 450 g * 4.18 J/g°C * 53.5°C = 100,507.5 J
This is the amount of heat energy that will be transferred to the water. To find out how much the temperature of the water will increase, we use the formula for ΔT again:
ΔT = Q / (mc)
Substituting the values for the water and the heat energy transferred:
ΔT = 100,507.5 J / (450 g * 4.18 J/g°C) = 50.5°C
So, the temperature of the water will increase by 50.5°C. Since the water was originally at 23°C, the highest temperature it will reach is:
23°C + 50.5°C = 73.5°C
So, the highest temperature the water will reach is 73.5°C.
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