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9. How much heat in Joules, must be added to 42.6 grams of water to increase the temperature of the water by 5.0^circ C ? (The specific heat of water is 4.18J/g^circ C) 890J 3.7J 149J 3720J

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4.7 (236 Votes)
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Raymond Veteran · Tutor for 11 years

Answer

890 J.

Explanation

This question uses the concept of specific heat capacity similar to the example. We're asked about heat in joules based on the temperature change, mass of water and specific heat capacity. Like the example, we use the formula q = mcΔT. Here, m is the mass of the water which equals to 42.6 grams, ΔT is the temperature change which is 5.0 °C and c is the specific heat capacity which equals 4.18 J/g°C referring to water. We substitute these values in the formula to determine the total heat (q) in joules. Thus, q = 42.6 g * 4.18 J/g°C * 5.0°C = 890.28 J approximately, a quantity of heat needed to raise temperature.