Question

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.

Answer 1

Answer:

T = 13.25°C

Explanation:

From the law of conservation of energy:

Heat Lost by Watermelon = Heat Gained by Cans

$m_wC_w\Delta T_w = m_cC_c\Delta T_c$

where,

$m_w$ = mass of watermelon = 6.48 kg

$m_c$ = mass of cans = (12)(0.35 kg) = 4.2 kg

$C_w$ = specific heat capacity of watermelon = 3800 J/kg.°C

$C_c$  = specific heat capacity of cans = 4200 J/kg.°C

$\Delta T_w$ = Change in Temprature of watermelon = 29.4°C - T

$\Delta T_c$ = Change in Temperature of cans = T - 3.78°C

T = final temperature = ?

Therefore,

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

T = 13.25°C