An object with a mass of $18 kg$ , temperature of $240 ^oC$ , and a specific heat of $6 J/(kg*K)$ is dropped into a container with $48 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2017-02-24T06:16:41

The water does not evaporate and the rise in temperature ot the water is $=0.13ºC$

Let $T=$ final temperature of the object and the water

For the cold water, $Delta T_w=T-0=T$

For the object $DeltaT_o=240-T$

$m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)$

$C_w=4186Jkg^-1K^-1$

$C_o=6Jkg^-1K^-1$

$m_0 C_o*(240-T) = m_w* 4186 *T$

$18*6*(240-T)=48*4186*T$

$240-T=(48*4186)/(108)*T$

$240-T=1860.4T$

$1861.4T=240$

$T=240/1861.4=0.13ºC$

The water does not evaporate

An object with a mass of 18 kg18 kg, temperature of 240 ^oC240 ^oC, and a specific heat of 6 J/(kg*K)6 J/(kg*K) is dropped into a container with 48 L 48 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?