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An ice chest at a beach party contains 12 cans of soda at 3.78 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 6.48-kg watermelon at 29.4 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.

Sagot :

hamzaahmeds

Answer:

T = 13.25°C

Explanation:

From the law of conservation of energy:

Heat Lost by Watermelon = Heat Gained by Cans

[tex]m_wC_w\Delta T_w = m_cC_c\Delta T_c[/tex]

where,

[tex]m_w[/tex] = mass of watermelon = 6.48 kg

[tex]m_c[/tex] = mass of cans = (12)(0.35 kg) = 4.2 kg

[tex]C_w[/tex] = specific heat capacity of watermelon = 3800 J/kg.°C

[tex]C_c[/tex]  = specific heat capacity of cans = 4200 J/kg.°C

[tex]\Delta T_w[/tex] = Change in Temprature of watermelon = 29.4°C - T

[tex]\Delta T_c[/tex] = Change in Temperature of cans = T - 3.78°C

T = final temperature = ?

Therefore,

[tex](4.2\ kg)(3800\ J/kg.^oC)(29.4^oC-T)=(6.48\ kg)(4200\ J/kg^oC)(T-3.78^oC)\\469224\ J-(15960\ J/^oC)T = (27216\ J/^oC)T-102876.48\ J\\469224\ J + 102876.48\ J = (27216\ J/^oC)T+(15960\ J/^oC)T\\\\T = \frac{572100.48\ J}{43176\ J/^oC}[/tex]

T = 13.25°C

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