An object with a mass of $3 kg$ , temperature of $270 ^oC$ , and a specific heat of $18 J/(kg*K)$ is dropped into a container with $36 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

Question

1275 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-20T14:04:16

The water does not evaporate. The increase in temperature of the water $=0.1ºC$

The heat transferred from the hot object, is equal to the heat absorbed by 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 $DeltaT_o=270-T$

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

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

$C_o=18Jkg^-1K^-1$

$m_0 C_o*(270-T) = m_w* 4186 *T$

$3*18*(270-T)=36*4186*T$

$270-T=(36*4186)/(3*18)*T$

$270-T=2790.7T$

$2791.7T=270$

$T=270/2791.7=0.1ºC$

An object with a mass of 3 kg3 kg, temperature of 270 ^oC270 ^oC, and a specific heat of 18 J/(kg*K)18 J/(kg*K) is dropped into a container with 36 L 36 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?