A 0.050kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing?
dH metal = - (dH water) m C dT metal = - (m CdT water) 0.050 kg (899J/kg*C) dT = - [(150 g 0 94.184 J/g-C) (25 C -21 C)] 44.95 dT metal = -2510.4 dT metal = -55.8C initial temp = 21.0 + 55.8 yoiur answer is 76.8 C
One liter of water at 30 degrees is mixed with one liter of water at 50 degrees. What is the temperature of the mixed water?
Let the first volume of water be [math]V_1[/math]. Let the second volume of water be [math]V_2[/math]. Let the initial temperature of the first volume of water be [math]\theta_1[/math]. Let the initial temperature of the second volume of water be [math]\theta_2[/math].Let the energy required to heat a unit volume of water by unit temperature by [math]C[/math].So, if the final temperature is [math]\theta_{final}[/math], we haveheat gained by the first volume of water[math]=V_1\times C\times (\theta_{final}-\theta_1)[/math]andheat lost by the second volume of water[math]=V_2\times C\times (\theta_2-\theta_{final})[/math].By the law of conservation of energy,heat lost = heat gained[math]\therefore V_1\times C\times (\theta_{final}-\theta_1)=V_2\times C\times (\theta_2-\theta_{final})[/math][math]\therefore V_1\theta_{final}-V_1\theta_1=V_2\theta_2-V_2\theta_{final}[/math][math]\therefore \theta_{final}=\dfrac{V_1\theta_1+V_2\theta_2}{V_1+V_2}[/math]Substituting the given values,[math]\theta_{final}=\dfrac{30+50}{2}=40^\circ C[/math].
A perfect gas at 27°C is heated at constant pressure till its volume is doubled. What will be its final temperature?
According to ideal gas equation PV=nRTP=pressure of gas,V=volume of gas=volume of container,T=temperaturueV1 =INITIAL VOLUME ,V2 =FINAL VOLUMET1 =INITIAL TEMPERATURE =27+273=300T2 =FINAL TEMPERATUREAccording to question,V2 =2 x V1Therefore,V1/V2 = T1/T2V1/2V1=300/T2T2 =FINAL TEMPERATURE=600KVidyanchal Academythank you
A piece of gold with a mass of 45.5g and a temperature of 80.5 degrees C is dropped into 192.0g of water at?
c(gold) * m(gold) * [ t(gold) - t(final) ] = c(water) * m(water) * [ t(final) - t(water) ] 0.13 * 45.5 * [ 80.5 - t(final) ] = 4.2 * 192 * [ t(final) - 15 ] 476 - 5.9t(final) =806.4t(final) - 12096 812.3t(final) = 12572 t(final) = 15.5 Therefore the final temperature is 15.5 C. Hope it helps!
A piece of copper metal is initially at 100º C. It is dropped into a coffee cup calorimeter containing 50.0 g?
A piece of copper metal is initially at 100º C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a temperature of 20.0º C. After stirring, the final temperature of both copper and water is 25.0º C. Assuming no heat losses, and that the specific heat (capacity) of water is 4.184 J(g•K), what is the heat capacity of the copper in J/K? a. 2.79 J/K b. 3.33 J/K c. 13.9 J/K d. 209 J/K e. none of the above
2 kg of ice at 0°c is mixed with 8 kg of water at 20°c. What is the final temperature?
Final temp. Will be water at 0 degree celcius
A 1.50 kg iron horseshoe initially at 580°C is dropped into a bucket containing 24.0 kg of water at 25.0°C. Wh?
A 1.50 kg iron horseshoe initially at 580°C is dropped into a bucket containing 24.0 kg of water at 25.0°C. What is the final temperature? (Ignore the heat capacity of the container, and assume that a negligible amount of water boils away.)