A jug containing 0.250 kg of liquid is put into a refrigerator. Its temperature decreased from 20°C to 15°C. The amount of energy transferred from the liquid was 5,250 J. Calculate the specific heat capacity of the liquid.
specific heat capacity = change in energy ÷ (mass × change in temperature) [1]= 5,250 ÷ (0.250 × 5) [1]= 4,200 J/kg°C [1]
BP
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