An object with a mass of $2 kg$ , temperature of $380 ^oC$ , and a specific heat of $25 J/(kg*K)$ 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 · 2017-03-08T15:12:26

The water does not evaporate and the change in temperature is $=0.09ºC$

The heat is transferred from the hot object to the cold water.

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

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

For the object $DeltaT_o=380-T$

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

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

$C_o=25Jkg^-1K^-1$

$2*25*(380-T)=48*4186*T$

$380-T=(48*4186)/(50)*T$

$380-T=4018.6T$

$4019.6T=380$

$T=380/4019.6=0.09ºC$

The water does not evaporate

An object with a mass of 2 kg2 kg, temperature of 380 ^oC380 ^oC, and a specific heat of 25 J/(kg*K)25 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?