# Q To Ml

Q To Ml – SciencePhysicsExpert solution / answer The specific heat of water is 4.18 kJ/kg. °C. The heat transfer expression for a cold drink is: Q = mc(T, -T) cold i cold 10 m (4.18 kJ/kg – ° C) (48 ° C -16 ° C) 1 ml = (120 ml) ) (10 J = (16.0×10* kJ) 1 kJ = 16 J I used the expert’s formula above to calculate (d), which is incorrect in part (c) but (d) is correct. .Please double check where the problem is. with the expert’s solution (c.) Solution (d) Q = MhotC (Tf – T;i heat) 16 J I used the expert’s formula above to calculate (d) This is a wrong answer to part (c) but (d) is correct… ..recheck where the problem is with the expert solution (c.) Solution (d) Q = MhotC (Tf – Ti heat) [ 10-5m²- (120ml) (4.18 kJ/kg.°C) (48-69) ° C) 1ml (103 (‘10.5 x 10 * kJ) () kJ = 10.5 Joules It is in the formula above. Worked in the part (d) … You don’t know why part c is wrong.

Expert solution/answer The specific heat of water is 4.18 kJ/kg. °C. The heat transfer expression for a cold drink is: Q = mc(T, -T) cold i cold 10 m (4.18 kJ/kg – ° C) (48 ° C -16 ° C) 1 ml = (120 ml) ) (10 J = (16.0×10* kJ) 1 kJ = 16 J I used the expert’s formula above to calculate (d), which is incorrect in part (c) but (d) is correct. .Please double check where the problem is. with the expert’s solution (c.) Solution (d) Q = MhotC (Tf – T;i heat) 16 J I used the expert’s formula above to calculate (d) This is a wrong answer to part (c) but (d) is correct… ..recheck where the problem is with the expert solution (c.) Solution (d) Q = MhotC (Tf – Ti heat) [ 10-5m²- (120ml) (4.18 kJ/kg.°C) (48-69) ° C) 1ml (103 (‘10.5 x 10 * kJ) () kJ = 10.5 Joules It is in the formula above. Worked in the part (d) … You don’t know why part c is wrong.

## Q To Ml

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(Q) What is a cold drink when you add 4 oz (120 ml) to a hot drink? Report your answer in Joules.

D) When you add 4 oz (120 ml) of cold drink, what is your calculated heat transfer (Q) to the hot drink? Report your answer in joules.

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Copy image text: Expert solution/answer The specific heat of water is 4.18 kJ/kg. °C. The heat transfer expression for a cold drink is: Q = mc(T, -T) cold i cold 10 m (4.18 kJ/kg – ° C) (48 ° C -16 ° C) 1 ml = (120 ml) ) (10 J = (16.0×10* kJ) 1 kJ = 16 J I used the expert’s formula above to calculate (d), which is incorrect in part (c) but (d) is correct. .Please double check where the problem is. with the expert’s solution (c.) Solution (d) Q = MhotC (Tf – T;i heat)

Copy image text: 16 J I used the expert formula above to calculate (d) and it shows that the answer to part (c) is wrong but (d) is correct… please check where it is .

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