Activity solve the problems below to practice what you have learned. Remember to use the formula for a heat transferred: Q=mC(T: - T:).
Part A
The specific heat of lead is 0.129 yg°C. Find the amount of heat transferred to lead when 355 grams are heated from 95°C to 140°C.
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2060.775 J

Explanation:

From the question given above, we obtained the following data:

Specific heat capacity (C) = 0.129 J/g°C

Mass (m) = 355 g

Initial temperature (Tᵢ) = 95 °C

Final temperature (Tբ) = 140 °C

Heat (Q) tranfered =?

We can calculate the amount of heat transferred by using the following formula:

Q = mC(Tբ – Tᵢ)

Q = 355 × 0.129 (140 – 95)

Q = 45.795 × 45

Q = 2060.775 J

Thus, 2060.775 J of energy was transferred.

c u c l 2 + 2 n a n o 3 → c u ( n o 3 ) 2 + 2 n a c l

  long weird answer but i hope this works