Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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 :

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