If heat can be transferred to a sample of aluminum at a rate of 0.260 J/s (assuming no loss of heat to the surroundings), how long will it take to raise the temperature of 64.8 g of the aluminum by 63.0 oC ?
belle clara daisy kate rose
.9000 is specific heat of Al J/gso (.260 J/s)(TIME) (.9000 J/g) (64.8g) Divide both sides by (.260 J/s) (.260 J/s)(TIME)/(.260 J/s) (.9000 J/g) (64.8g)/(.260 J/s) so you get Time (.9 J/g)(64.8)/(.260 J/s) Time equals 224.31 seconds so you have to round off getting 224 seconds or 3 minutes and 44 seconds
belle clara daisy kate rose
.9000 is specific heat of Al J/gso (.260 J/s)(TIME) (.9000 J/g) (64.8g) Divide both sides by (.260 J/s) (.260 J/s)(TIME)/(.260 J/s) (.9000 J/g) (64.8g)/(.260 J/s) so you get Time (.9 J/g)(64.8)/(.260 J/s) Time equals 224.31 seconds so you have to round off getting 224 seconds or 3 minutes and 44 seconds
