A 400W immersion heater is used to heat a liquid of mass 0.5kg. lf the temperature of the liquid increases by 2.5oC in one second, calculate the specific he...
Question 1 Report
A 400W immersion heater is used to heat a liquid of mass 0.5kg. lf the temperature of the liquid increases by 2.5oC in one second, calculate the specific heat capacity of the liquid (Neglect heat losses to the surroundings)
Answer Details
The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of 1 kg of the substance by 1 Kelvin or 1 degree Celsius. In this problem, we have an immersion heater that provides 400 W of power to a liquid with a mass of 0.5 kg. The temperature of the liquid increases by 2.5oC in one second. We can use the formula: Heat energy supplied = mass x specific heat capacity x change in temperature to calculate the specific heat capacity of the liquid. First, we need to calculate the heat energy supplied by the immersion heater in one second. We can use the formula: Power = Heat energy supplied / time Rearranging this formula, we get: Heat energy supplied = Power x time Substituting the given values, we get: Heat energy supplied = 400 W x 1 s = 400 J Next, we can use the formula for specific heat capacity to solve for it: specific heat capacity = Heat energy supplied / (mass x change in temperature) Substituting the given values, we get: specific heat capacity = 400 J / (0.5 kg x 2.5oC) = 320 Jkg-1K-1 Therefore, the specific heat capacity of the liquid is 320 Jkg-1K-1. The answer is.
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