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The specific heat of copper is 385 J/kg·K. If a 2.6 kg block of copper is heated from 340 K to 450 K, how much thermal energy is absorbed?

2 Answers

5 votes
Answer: 110,110J or 110kJ

Q=mc*change in temp

Q=(2.6kg)(385J/kg*K)(450K-340K)=110,110J

Q=110kJ
User Nicolas Siver
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6.2k points
2 votes

Answer:

Heat absorbed, Q = 110110 J

Step-by-step explanation:

It is given that,

The specific heat of copper is,
c=385\ J/kg.K

Mass of block, m = 2.6 kg

Initial temperature,
T_1=340\ K

Final temperature,
T_2=450\ K

Thermal energy is given by :


Q=mc\Delta T


Q=mc(T_2-T_1)


Q=2.6* 385* (450-340)

Q = 110110 J

So, the thermal heat of 110110 J is absorbed. Hence, this is the required solution.

User Levous
by
6.1k points