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A liquid of specific heat 3000J/kgk rise from 15°c to 65°c in 1 min when an electric heater is used. If the heater generate 63000J, calculate the mass of the water​

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Answer:

0.42 kg

Step-by-step explanation:

Heat is proportional to mass by way of the conversion factor that is the inverse of the specific heat.


\frac{63000\text{ J}}{\frac{3000\text{ J}}{\text{kg$\cdot$K}}\cdot(65-15)\text{ K}}=0.42\text{ kg}

The mass of the liquid is about 0.42 kg.

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